iLIBFiARY OF CONGRESS. I 

f # 

t^^^ I^^^^Jfc'^' I 



! UNITED STATES (JV 'AMfRfcA. f 



Copyright: 
By GEORGE I. ALDEN. 



NOTES 



Rankine's Applied Mechanics. 



GEORGE I. KlDEN, B. S., 

PROFESSOR OF THEORETICAL AND APPLIED MECHANICS 

IN THE WORCESTER FREE INSTITUTE, 

WORCESTER, MASS. 






q9oM'' 



HARTFORD, CONN.: 
Pbess op the Case, Lockwood & Beainabd Company. 

1877. 






^-'^lioi 



rCsTTEODUCTIOlSr. 



The following pages are the result of putting in permanent 
lorm some of the matter which it has been found expedient or 
necessary to give by dictation to students in the Worcester 
J^ree Institute, who pursue for the first time, the study of Ran- 
kine's Apphed Mechanics. The object in their publication is not 
to turnish a key, or provide a substitute for diligent study and 
careful thought on the part of the student, but rather to en- 
courage him by giving such suggestions, solutions, and refer- 
ences as experience has shown that the average student requires- 
thus economizing time in the preparation of the lesson, and also 
giving the instructor opportunity to devote the time spent in 
the class room to recitations, and to the application of the prin- 
ciples and formulae of the lesson, to practical problems. 

To what may be strictly called notes on the "Applied Me- 
chanics," I have added a brief explanation and illustration of the 
method of producing the reciprocal diagram of stresses sub- 
stantially taken from "Economics of Construction," by R. E. 
Bow, C. E. Also a separate treatise on strength of beams and 
an investigation of a particular problem relating to seven bar 
parallel mot;ions, known as " Peaucillier's Parallel Motion." 

This work has been prepared from materials drawn from 
various sources, especially from notes given by Prof Eustis of 
ihe Lawrence Scientific School, to the class of '68. 

I have also received assistance from George H White B S 
a graduate of the Free Institute, and have inserted on several 
articles of the Applied Mechanics, notes which are entirely his 
own work. I have endeavored to make proper reference to 
works from which quotations or extracts have been taken 

ihe blank pages at the end are intended to receive such sup- 
plementary notes as the instructor may find adapted to the 
needs and capacity of his class. 

w ^ r GEORGE I. ALDEN. 

WOKCESTER Free Institute, Feb. 1st, 1877. 



INTEGEATIOK 



The following integrals are of frequent occurrence, and are 
here given for future reference : 

(A) fd X Va' - x' 

In the general formula for integrating by parts, I udv= ■ 

u V — I V . d u, let u — \/a^ — x^, and d x = d v ; then 

X d X 
d u =1 — — _ and x ^= v 

d X ^Ja^ — X- — X ^/ d^ — x^ A- / , 

^ J V a?- x" 



r , nd X (a^ — x'^) n x^ d x 

.: dx ^0} -x' = — ^ ^ = - , + 

'^ J ^/ d' - x' J ^/o? -x' 

/a? d X 
^fcF^^x"- 

/. , /» d^ d x 

d X Vtt' — x"" = X Va' = x' + / — = 



Va' - x' 

d i-\ 

/d^ d X \ a / X 

, =: d' , = a' sin-1 — + C. 

Va' - x' a/1 - x' <^ 

■^ 

-•. 2 Cd X V«' - x^ = X Va' - x'' + d" sin-i — + C 

.-. fd X ^/a' - x' = — Va^ - x' + — sin-^ — + C. 

/d X 

Let Vd' -f x^ = z — x: Then d -{■ x^ =z z^ — 2z x -\- ^ ox 



INTEGRATION. 



a} ^^ z^ — 1 z X. The differential of this equation is = 1 z 

(z — x) d z 

a z — 2 z a X — 2 x a z .-. a x :=i 

z 

/d X (z — x) d z P d z 
, = -, A = / = log, ;2 + C = 

iog,(x+v^^r:p^) + c 

(C) Jd X ^a' + x' 

In the formula for integration by parts, I u d v = u v — 

I V d u, let '^d^ -f x^ = u and d x = d v. Then x = v and 

X d X 
—j===z = d u 
Vo? 4- x' 

d X Vd' -Ux" = x Va' J^ x' — / , . Again, 

J ^ J V a' -\-x' J Va' -I- ;r^ ^ 

/x^ d x 
Va' -f x' 

.-. 2 /'f^ X Va'' 4- x' = X V^2-Tr^ i ^2 /" /^ 
t/ ^ ^ ^ t/ 1/a2_^cc^ 

= =: a^ logg 

(;r + Va' ^ x') + C. • 

.-. fd X Va' J^x'= -^ Va' + x'^+-^, log,(a:+Va2-|-x'^)4-a 

(D) • A^ c^ X Vd' — x' 

In the formula udv = uv — / v du, let x dx Vd^ _ x^ 

1 3 

= cZ 7;, and X = u : then v = _ - (a^ _ x^) and d u = d x 

.-. A^ c^ a; Va2 _ ^2 ^ _ ^ j^,,2 _ a;2>)i _|_ 1 A? ^ (^2 _ ^2^1 
= - — (a' - xy 4- I fd X {a' - x') V^^^^T^ 



INTEGRATION. 



= -^ («2 _ xy + i CdxCL'Vd' - X^ 

— - I x^ d X Va^ — x^ 
3 J 

Transposing the last term of the last member, we have 
- fx" dx Va? — x' = - — (a^ _ x')^ + - fa' dxVa'-x^ 
From integral A we have 

la' fdx viF:r^'^'^\~viF^r^-j^^ sin"-'— l + c. 

3t/ o \ 2i 2 a ) 

.-. ^ fx' d X V^^ZT^- = _ |- («2 _ ^2>)l _|_ _?^ V^TZ"^ 
a* , -I X ^, 

■■.A 



' f? a; Va'^ - x' = - ^ (a^ - x^)^ + i^ Va' 
4 8 

sm — 4- C. 



This integral, taken between the limits a and 0, reduces to 
— sin — , which can easily be memorized. 

(E) /*cos2 Odd. 

Integrals of the powers of sin d and cos are found by sub* 
stituting for these functions of 6, their values in terms of the 
multiple angles, as in the following case : 

r . . /'I + cos 2 , sin 2 * 

J cos^ ddB^ J -^-^ ^ ^ == T + —4— + C. 

Article 83. * 

In solving the following problems the student should always 
sketch a figure representing the surface or solid under consid- 
eration, and one of the elementary parts into which it is conceived 
to be divided, and determine the limits for the integral by an 
inspection of this figure. In double or triple integrations, we 

NoTR —The fifPt three integrals are taken from Todhnnter''s Intecrral Calculns. 
Note.— The abbreviation A. M. will be used for Eankiue's Applied Mechauics. 
* The reference is to Article 83 of the Applied Mechanics. 



ABTICLE 83. 7 

must, in general, first integrate with respect to one of the vari- 
ables, behueen the proper limits, and express the result in terms of 
the other variables. This may be continued until the complete 
integral is obtained. 

To illustrate this process take the general formulae for center 
of gravity of a solid, viz: 



/ / I X dx dy dz 
III dx dy dz 



and similar values for y^ and z^. 



The following application of these formulae is found in Tod- 
hunter's Analytical Statics. 

Problem. Find the center of gravity of the eighth part of an 
ellipsoid cut off by the three principal planes. 

Let Fig. (1) represent the solid in q;uestion, the equation of 

the surface being —- + — - -f -— = 1. (1) 



If we integrate 
first with respect 
to 2, between the 
limits z, and zerd, 
we include all the 
elements {dx dy dz) 
in the prism P Q. 
Next integrate 
with respect to y 
between the limits 
2/i and zero. We 
thus include all 
the prisms in the 
slice between the 
planes ZL and 
m M. 

F ro m Equa- 
tion (1), 2i =zz c 




y, = h\/l-. 



~ ; from Equation of ellipse in the plane X Y, 



AETICLE 83. 



Thus a7o 



/ I I X dx dy dz j j x dx dy z^ 
I I dx dy dz dx dy z 

r r c dx dy A/^_^_f r r \ dx dy 

O o tJ o f d^ y^ d o d o b 






The other coordinates of the center of gravity will easily be 
found by taking moments with reference to the axes OX and 
OZ, and following the integration as above indicated. 

Problem III. Draw a line through C parallel to E B, form- 
ing a parallelogram and a triangle. A line Joining C and the 
center of gravity of the triangle will be parallel to D. 

Let / B9D = d. Then equating the momer^t of the trapezoid, 
about A B, to the moment of this parallelogram and triangle, 
about the same line, we have 

— — — . D. sin Q . X, sine .w =z h 01) sin . sin 6 . iv 4- 

2 ° 2 

,^ ,, "OU . ^ OlT . ^ B + 5 OD/B-Z* , , \ 

(B-5)-^sm9.-^sm0...; -f~ • ^o = -^(^- + ^J 

_ OD" /2 (B-5)J-6 l\ _ ~0^ /2 B 4- 4 JV _ OT 

•*• ^» ~ ~^r \ 3 (B+6) ; ~ ~ir \ 3 (b+&) / ~ ~2~ 

/ 3 (B+5) - (B-5) v _ "Op- / B - 5 V 

\ 3(B + ^) / 2 \ 3(B+Z.)/* 

Problem V. Taking moments about Y we have for the 
arm of the element ydx, the distance x . sin Y X. The dis- 
tance of the center of gravity from Y on a line at right angles 
to Y is aro • sin Y X 

I X y dx I xy dx 

.•.x„ sin YOX=siay 0X^^-1^ '''0 = ^, 

I y dx I \j dx 



2/o = 



ARTICLE 83. 9 

j ' lyy dx I y"^ dx 
y dx 2 / y dx 

o do 

The equation of tlie curve is 3/=^ = 2>' x. 

Problem VII. In this, as in many problems relating to the 
circle, it is convenient to use polar coordinates. Let (Fig. 32 
A. M.) be the pole, p the radius vector, and the variable angle 
measured from the axis X. The circle is divided into differ- 
ential areas, by concentric circles d^ apart, and by radii making 
an angle d(\> with each other. 



d 



9- 



Its distance from Y = p cos 0. 
Its moment about Y = w p^ d p . cos (p d (p. 
. in this case 

'^ I / P ■ P ■ ^^^ ({> ' d(f) "o" / cos (j) c 



Vo-- 



_ 2_ ^ (sin0)+'^^ =1.R / sin0-sin(-(^) \ 2 ^ sin_0 

/xy dx j x{i''^—x^')'^dx 
r cos % t/ r cos Q 
rROBLEM V ill. X^— -^, = — — 

/ y d X I {f — x^)2 d x 

t/ r cos J r cos 

r r y dy dx I - (r"^ — x^) d x 
^JuJ f cos _ t/ r cos 2 

r r dydx r {r^ ~ x^)^ d x 

J ot/ r cos Q ' tJ r cos Q 

Problem XVI. Take moments about Y. The moment of 
the annular wedge is equal to the moment of the whole wedge, 
minus the moment of the interior wedge. (See Art. 76 A. M.) 

Compare with Problem 15. 

Problems XIX and XX. Conceive a plane passing through 
X, (Fig. 39, A. M.) making a variable angle y with the plane 
X Y. Let p be a radius vector in this plane, making a va: iable 
angle <^ with X. 

Then pdcf) dp is an elementary area in the above described 
plane. "When th^t plane revolves about X, through the angle 
2 



10 ARTICLE 83. 

dy, tMs elementary area describes a volume which is measured 
by the area times the distance through which it moves. This 
distance is the arc of a circle whose radius is p sin '0, and angle 
dy, and = p sin ^dy. 

.'. Elementary volume = pdcp . dp . p sin (pdy. 

Its distance from the plane Y Z = p cos 0; Z being the 
third rectangular axis. 

/ / / p d (j) . d p . p sin (j) dy . p cos . lu 



///' 



p d ({) . d p . p sin ([) dy . w 

I j / p^ d p. sin (f) cos (p d (j). dy 

I I / p^ d p. sin (j) d (f). d y 

Integrating between proper limits, this formula will give the 
value of Xq for these, and similar problems. 

) r ^ /3 ) 27r 

In problem XIX the limits are p [- ; r ; 7 [ 

" " XX " " pV ^i?; yi"^^ 

In problem XX, and in similar problems, 

/ / / pd(p . d p . psin(f)dy . p sin cos y . w 

/ / / p d (p . d p . psin(f) d y . w 

The limits for this integral, in problem XX, are as stated 
above. 

Article 91. 

The expressions for moments around X, Y, and Z, on 

page 74, are easily obtained by resolving the force P, having 
its point of application in the plane X Y, into components par- 
allel to the three axes, and taking the moments of these com- 
ponents. Equation 4 is reduced by the relation cos^ a -\- cos^ /3 
-{- cos^ y = 1. See Church's Analytical Geometry, Art. 48, 
Equa. 4. 

Equation 5 A has for its first member the sum of the rectan- 
gles of the cosines of the angles which two lines make with 
three rectangular axes, which is equal to the cosine of the angle 
between the two lines. See Church's Analytical Geometry, 
Art. 48, Equa. 5. If this cosine is 0, the angle must be 90°. 
The equation is verified by multiplying the first of equations 5 



ARTICLE 95. 11 

by cos a, the second by cos /3, and the third by cos y, and add- 
ing the products. The second member of this sum must then 
be reduced by introducing the values of M^ Mg, M3, from Equa- 
tion 3. 

Aeticle 95. 

Equations 3. — The point whose coordinates are expressed by 
this equation, is situated in the angle Y' X^ 

Problem. 

Find the moment of inertia of a rectangle of length h and 
breadth J, about a neutral axis perpendicular to its plane. See 
Equation 5. 

Ans. -y- (j'+A^)- 

Equations 13. — Complete figure 45, A. M., by drawing a line 
from C, parallel to T, and call the point where this line inter- 
sects the hne Xj, N. From N, draw a line parallel to T C, 
and call its point of intersection with T, H. The angle 
T Xi = 90° - / Xi Y^ = p. 

Take the most general form of the equation of a straight 

A B 

line, Aa;-f-B?/-[-C=:0, and write it in the form tt ^ + tt 

y -\-\ — 0, Since this equation is true for all values of x and y, 
it will be true when x or y becomes 0. 

Let the intercept on the axis X. = c, and the intercept on 
Y = t^, so that, when y — 0, x = c, and when x =z 0, y = d. 

A A 1 

Make y = 0. Then -^c4-l = 0, or-7^= 

O L> c 

■p Ti 1 
Make x = 0. Then -^ c^ + 1 = 0, or -7^ = -- 

These values introduced into the general equation, reduce it 

X y 

to 4. _|_ 1 - 0. 

c a 

or - + ^=.1. (A) 

Now multiply both members of this equation by n, drawn 
perpendicular to the straight line, and making the angle /3^ with 
the axis O X. See line T, fig. 45, A. M. 



12 AKTICLE 95. 

^,71 n ^ n , , ^ 

Then — x -\- -j y =n. But — = cos /3^ and —r = sm/3^ 

.'. X COS /3^ + 2/ sin /3^ = n. (B) 

The equation of a tangent to an ellipse (T C) is a^ y y^ ■\-l'^ x 
x^ = a^ y^ (see OIney's General Geometry, Art. 136) 

X 33^ V V^ 

— ^ 1 p- = 1. Multiply this equation by n. 

^ —^y^-=n. (C) 

By comparing equations C and B, it will be seen that cos /3^ 

^^ n ^ , ^^ y^ n 
= -^, and sm /3^ = -^ 

^ a' cos /3^ . ^ 6^ . sin /3^ .^. 

.♦. 05^ = !— , and V = — ' (-D) 

Substitute these values in equation B, for x and y, 

^l-^I^ '^JLfL =n- n' = a' cos^ /3^ + Jl sin^/B^E) 

To find the value of n t. 

The general equation of a normal to an ellipse is y — 3/^ = 

a^ y^ 

-7^— j-(a: — £c^). (See OIney's General Geometry, Art. 156). 

In this equation make y = 0, and divide both members by y^. 

a" 
— 1 = 72 1 (^ — x^); .*. — b"^ x^ = a^ X — a^ x^; x => 

K-f)-'loK. 

al cos /3* 



From Equation D, we have a;^ = 



n 



(g^ - ¥)cos(3\ 

(a^ - 52) cos /3^ . sin /3^ 



N. sin /3' = N H = C T = i! = ■ 

.'.nt= (a^ — ¥) cos /3^ . sin /3^ 

Problem IY. 

Use polar coordinates. An elementary area = p d (f> . dp. 
(See Note on Problem VII of Art. 83.) Its distance from 



AKTICLES 102-112. 13 

Y = p cos (f). Its moment of inertia = p d cp d p {p cos cff 

gral E). 

Article 102. 

Equations 1 and 2 are given in Chauvenet's Trigonometry, 
page 163, formula 53, and page 162, formula 48. 

Article 105. 

Problem II. — According to the preceding notation, tlie form- 
ulae near the top of page 91 A. M. should be 
D = ^^^ . area B C + ^j,^ . area C A + p^^ . area A B. 
E = j9^ y . area OB C -{- p^^ . area C A -{- ^^^ . area A B. 
0¥ ==p^^. area B C -f i^y^ • area OCA ^-Pzz • area A B. 

Articles 110 and 111. 

The student should carefully study these articles, and mem- 
orize the theorems, in order to understand Art. 112. 

Article 112. 
Equation 1. — From figure 54, A. M., we have 



OR^ = OM^ + MR^4-20M.MR.cosRMN 

or;,/= {?^y + (?^y + ,(?l^) C0S2 A 

pi + Pv P^ - Pv A JO 2 A 

_ ^ ^y _i^ ^' ^ i'v COS 2 a; 7i = ^ (1 + cos 2 a; ^) 

»2 A 

+ ^ (1 - cos 2 (c n). 

A • A A 

But cos 2 a; 71 = 2 cos^ ccti— 1 = 1 — 2 sin^ x n. 
A A 

... p^ = p^ cos^ ^n -\' p^ sm^ X n 

J\ A A ) 

.'. p^—y \p^ . cos^ xn'\- p^ sin2 x n\ = '0 R. (1) 

Equation 2.— From the triangle M R, we have 
sin N R : sin R M : : JifR : 0^ 



14 ARTICLE 112. 

or sin N R : sin (180 - 2 x n) \\ ^^~^y ; p^ 

A A ^ _„ 
.-. sin N R = sin 7^ r == sin 2 a; ^ -^ ~ 

A / A ^ _p\ 

or N' R = ^ r = arc . sin 8 sin 2 a; % ^- 1 • (2) 

Proof that the locus of the point R in an ellipse whose semi- 
axes are p^ and py. 

Equation 1 squared, becomes 

A A A A 

pi = pi cos^ xn -j- pI sin^ xn =z pi— pi sin^ xn -|- pi sin'' xn 

. A . A 

.'.pi = pi. sm^ icTi •\- Pl — pI sin'^ a;w. 

A A 

pi pI = pi pI sin^ xn -\- plpl — Py sin^ xn. 






^ /\. Multiply and divide by /)^. 

pi pI sin^ a;^ -\- pIpI— Pt ^^^^ ^^ 



^'^ ~ A A 

PIpI sm^ ic^ ^2^2 _ ^4 31^2 a;^ • / a\ 

— ^^ + ^ (^) 

Call the angle FOR, a\ From the triangle FOR we have 
sin FOR : PR : : sin OPR : OR; PR = MP - MR = OM-RM 

_ P. +Py Px-Py _ „ 
~ 2 2 ~'^^' 

Angle OPR = /POM = xn, smceMP = OM. 

. A 
, -A . , p,,.sin xn 
.'. sm a : PyW sm xn : p^. or sm a = tl. . 

Pr 

A • -A 

... pI sin^ xn „ , , . . , pl — pl - sm=^ xn 

sm'^ a' = t-^ ; cos^ a' = 1 — sm'^ a' = t-^^ ^-^ 

pl Pr 

A A 

pl pl sin^ ccTi ^ ' r pIpI— Pt sm'^ a;?i , o / 

.'.■L^^±y = pl sm^ a'; iX^i — ±^^^ = pl cos^ a'. 

Hence equation A becomes, 



2 /y)2 



^^ 



_ K P'l 



' pl^m^ a' -\-fyQO^^a' 



ARTICLE 112. 15 

But sin a' = -^ = — : cos a' = --— = 

OR p/ OR p^ •" 

•'^r 2 2 ; ^^2 2^2 _|_^2 ^2 _ ^2 ^2 _ tJ^g 

equation of an ellipse whose semi-axes are p^ and Py. 

Equations 3 and 4. 
p„ = OM - MR cos (180° - 2 ;^) = ^^:^+^^Z:^ cos 2 xn 

p /\ 7) A A 

== ±-^ (1. -f COS 2 xw) + ^ (1 — cos 2 xn) =^p^ cos^ ccti + 

A 
Py sin^ xn. 

,^^ .A V —V .A A . v.A 

^j z=r MR sm 2 xTi = -L^ — :^ 2 sm a;w cos xti = {p^ — Py) sxn xti 

A ^ 

COS ccw. 

Comparison of figures 54 and 57, A. M. 

In figure 54 suppose a second plane making an angle with 
the plane A B, to be drawn through 0. 

Let the same construction be made to represent the compo- 
nents of the principal stresses on this plane, as is made for AB. 

The distance set off on the normal to this supposed plane, 
must be equal to OM, and the line corresponding to MR, must 
be equal to MR. There will also be a line corresponding to R. 
Now suppose this second system of lines to be revolved about 0, 
until the plane coincides with AB. The result will be ORMR'O 
of figure 57. By the construction of figure 57 it will be seen 
that MR =: MR' and OM is the same for both systems of Hues, 
hence the figure satisfies the condition of the problem, viz : that 

OM =^P±±Iy and MR = E^lZL^i . 
2 2 

Equation 15. — From the triangles OMR and OMR' figure 57. 

MR2 _ Qyi2 4- ^2 _ 2 pOM cos wr. 

MR'2 ^ 0^2 _|_ y2 _ 2 y OM cos ?//. 

,•. = _p^ — p'^ — 2. OM (p. cos nr — p' cos nV). 
p^ — p''^ 

... OM ■= — 7^ jr ' (15) 

2 (^. cos nr — p' cos ?iV) 



16 ARTICLE 112. 

^ Equations 16, are found by substituting - ^^ ~ri^y f^j, Qjyf ^j^ 

the above values for MR, and MR'. 

Equations 17. — For the first of these equations, draw a line 
from R perpendicular to ON (Fig. 57) and call its intersection 
with ON, A. 

Then cos 2 :^ = cos NMR = ^^. MR =?^:zh. 

MR 2 

MA = OA - OM = OR cosNOR - OM =^9 cos nr -^^-^ 

A 

2p cos nr— p^ — Py 

_ _ 

A 
A 2p . cos nr — p^— py 



.'. COS 2 xn = ■ 



The value of cos 2 xn^ is found by drawing a perpendicular to 
ON, from R', and repeating the above process. 

Equation 27. 

From equation 25, p + p' =: V^- «» «'"■ . (A) 

cos (p 

Write equation 25 in the form 

A 
cos'^ 6 (p"^ -\- 2 p p^ -}- p'^) =i4.pp' cos'-^ nr, suotract i pp^ cos'^ <p 
from both members of the equation, and extract the square 
root. 

Then p —p' ^=^1 ^/p p'. V cos^ nr — cos^ (^ (B) 

cos (^ 
From equations A and B, we find 

/ ^ ^ NX 

y p p' ( cos nr -\- v{ cos^ nr — cos^ (f)) 

cos (p 

I —i/pP' (cosnr — v(cos^ nr — cos'^ cf)) 



cos 

^f cos nr — v( cos'^ nr — cos'^ <^ 
P cos nr -j- v( cos^ nr — cos^ (f) 



(27) 



ARTICLE 124. 17 

Article 124. 

Formula for the center of pressure of a fluid, upon a plane surface. 

In figure 62, let BF be the plane and suppose FO to be a line 
representing the continuation of this plane to 0, the surface of 
the fluid. Let OB be the axis of a:, and the axis of y at right 
angles to it. 

Let the angle that OB makes with the surface of the fluid be 
represented by Q. 

Let F be the origin of coordinates, and x be measured pos- 
itively downward. 

dxdy = differential portion of this plane. 

The pressure on a unit's surface of this plane is normal to 
the surface, and measured by j9 = iv (x -f- OF) sin 6, in which 
lu is the weight of a unit volume of the fluid. (See Art. 110, 
A. M.) 

Let OF = a. Then p z= w (a -\- x) sin d. 

By formula 2 of Art. 89, A. M., 

/-\-y />FB 
/ w X (a -\- x) sin 6 . dxdy 
—yUo 



«/ —y Jo 



2/o = 



ID (a -|- x) sin d dxdy 

/-l-y />PB 
/ (^ax -j- x^) dxdy 

/ a -\- x) dxdy 

/y /»PB 
/ dxdy y {a ^ x) 
. —yt/o 

/y /^FB 

/ dxdy {a -\- x) 



(A) 



If the plane surface is symmetrical with respect to the axis cc, 
the above equations become 

/FB 
» y {ax -j- a;^) dx 

X, =: -7^ ■ —- • 2/o = 0- (B) 

/ y {a -\- x)dx 

If the plane extends to the surface of the fluid, a becomes zero, 

/ a? ydx 

and the equations B reduce to x^ = •— | , h being the 

/ xydx 

Jo 

depth of the plane surface. (C) 

3 



18 ARTICLES 124-159. 

Problem 1. — Find the position of the center of pressure on 
a rectangular plane surface whose depth is 7i, and breadth h, 
when its upper edge is parallel to the surface of the fluid, and 
at a distance a below it. 

From equation B we have 

I y {a X -\- x^) d X ^ 

I y (a -\- x^ d X 

r^ f , ■j\ J i a x^ x^\^ 

•■•^"~ /•'• ,, ~/ a;-^V''- 6a A + 3 A^' 

J ^(a + x)dx \ax +-;^ 

• Problem 2. — Find the centers of pressure of the following 
surfaces, when their bases are in the surface of the fluid. (The 
depth of each surface = h.) Rectangle, triangle, and parabola. 

2 14 

Ans. a^o = o h 9 h H h respectively. 

Problem 3. — The parabola with its vertex in the surface, and 
base parallel to it. 

5 
Ans. Xq = ^ h. 

Problem 4. — Find the center of pressure on a triangular sur-_ 
face whose depth is h, base parallel to the surface of the fluid, 
vertex upward and at a distance a below the surface of the fluid, 

4a7i2_j_3A3 
Ans. Xo — -?^ — -^ — — T-TT = distance froni vertex. 
6 a /i -|- 4 /r 

Article 159. 

VJ 

Equations 4. — The load — at 5, is resolved into two compo- 

w 
nents,. — cos i, which is transmitted through 5 4 and produces 

a stress on all the pieces of the secondary truss 15 3 4, and 

w 

— sin t, one-half of which is supposed to cause a pull on the 

piece 1 5, and one-half a thrust on the piece 5 3. 

Equations 4 are obtained by computing the stresses due to a 

Note.— It will be noticed that xq is measured downward from the top of the plane, 
and not from the surface of the fluid. 



ARTICLE 159.- 19 

w 
force — cos i acting in the direction 54, and combining these 

w 
stresses with the component — - sin i above mentioned. 

w 
Draw a hne parallel to 54 and equal to — - cos i. From its 

extremities, draw lines parallel to 14 and 34, and from their 
intersection draw a line parallel to 13, to meet the first line, to 
which it is perpendicular. 

w w 

Then from this diagram R43 = K^^ — -— cos «, cosec « = — - 

cotan i. 

R35 r= — cos i cotan i (from diagram) -|- - iy sin i (see above) 

tu /cos^ i 4- sin^ i\ w 

=Z -— \ r-^ r- I = —- cosec I. 

8 \ sm ^ / 8 

w- . ..." 

R51 z=:i -—- cos ^ cotan i (from diagram) — ^ '^^ sin i (see above) 

ty /•! — sin^ ^ — sin^ i\ lu /I — 2 sin^ A w 

^^ ~8 \ sirTT / "^ "8" \ suTT / "^ ~8 

(cosec ^ — 2 sin ^). 

TAe Method of Drawing Reciprocal Diagrams. 

Since the forces acting through any joint of a truss must be 
in equilibrium, these forces may be represented by a closed 
polygon whose sides are parallel to their directions. 

A complete diagram for a truss under a given load ought, 
therefore, to contain as many such polygons as there are joints 
in the structure. In most cases a variety of diagrams can be 
drawn, any one of which will correctly represent the stresses in 
a given truss. The one that fulfills most nearly the condition, 
that for each joint in the truss, there shall be in the diagram 
a closed polygon whose sides are parallel to the forces acting 
through that joint, and tal-en in tlie same order as those forces^ 
has been called the reciprocal diagram. 

A method of notation given by R. H. Bow, C. E,, renders the 
construction of reciprocal diagrams very easy in most cases. 
This method consists in placing upon the figure of the truss, a 
letter in each of the angular spaces formed by the intersection 
of the forces at each joint (a single letter in an enclosed area 
answering for all the internal angles formed by its sides), and 



20 



EECIPROCAL DIAGRAMS. 



in the diagram tlie same letter at the junction of two lines rep- 
resenting concurring forces, that is found in the angle between 
those forces upon the figure of the truss. 

To illustrate the application of this notation, let us construct 
the diagram for the frame represented by Fig. 18 A. M., on 
the supposition that it is loaded with W" uniformly distributed 
over 12 and 13. 

1st. Distribute the load upon the joints, as directed in- Art. 
147 A. M., or for a uniformly distributed load by placing upon 
each joint half of the load between the adjacent joints. This 

distribution gives - W at the joints 4. 6, 1, 8 and 10, and — "W 



at the joints 2 and 3. 




2d. Represent the directions of 
the external forces by lines, as in 
the accompanying Fig. 1. The 
forces are all given in direction, 
the internal forces or stresses be- 
ing represented in direction by 
the pieces of the frame. 

3d. Letter the angular spaces 
formed by concurring forces. 

In Fig. 1, P occupies 
the space between 35 
and the left hand sup- 
port, 3 5 and 5 7, 5 7 and 
7 9, 7 9 and 9 2, and 9 2 
and the right hand sup- 
port. K occupies the 
angular space at the left 
of the left hand support, 
between it and the load 
at 3. A occupies the 
angular space between 
3 5 and 3 4, 3 4 and 4 5, 
and 4 5 and 5 3, and so on. 
4th. Draw a line of 
loads (vertical in this 
case) representing the 
whole load W, and di- 
vide it into parts repre- 
senting the external 
forces acting at the 
joints. 



EECIPROCAL DIAGRAMS. 21 

As the supporting forces here are vertical and equal, they 
will each be represented by half this line of loads (K P and 
L P, Fig. 2). 

5th. Letter the line of loads to correspond with the lettering 
of the truss. In reading from K to a, Fig. 1, we cross the Hne 

"W 
representing the load ^^ at 3. Therefore K a on the line of 

w w 

loads, must be ^9 * ^or a similar reason a h = --, and so on. 

Since the loads at 2 and 3 rest directly and vertically upon 
the supports of the truss (and the supporting forces are vertical), 
they produce no stress on the frame, and may be left out of 
account. 

Then a h upon the line of loads will represent the effective 
load, and a P and h P the supporting forces to sustain that load. 
The loads at 2 and 3, and the letters K and L could, therefore, 
be omitted in Figs. 1 and 2. 

6th. Commence at some joint where all but two of the forces 
are known, draw the polygon of forces for that joint, and by 
proceeding from one joint to another complete the diagram. 

Commencing at the left hand joint of the truss Pig. 1, and 

"W 
leaving out the load — as above directed, we have on the line 

of loads, Fig. 2, the distance a P, upon which to construct a tri- 
angle having its sides parallel to 3 4 and 3 5. 

Draw through a, Fig. 2, a line parallel to 3 4, Fig. 1, and 
through P a line parallel to 3 5, and mark their intersection by 
the letter A. 

Then pass to the joint at 4, draw lines through A and h of 
Fig. 2, parallel to 4 5 and 4 6 of Fig. 1, and mark their inter- 
section with B. Through B draw a line parallel to 5 6 to meet 
a line through P, parallel to 5 7, and mark the intersection C. In 
a similar manner complete the diagram represented by Fig. 2. 

The advantage of this notation lies in the fact that, if the let- 
ters in the angular spaces about any joint are taken in order, 
and the same letters, taken in the same order, are found in the 
diagram, at the vertices of the polygon representing the forces 
at that joint, the diagram is correct. Thus in Fig. 1, for the 
joint 3, we have aAPa, and in Fig. 2, aAFa forms a closed 
polygon. Reading about joint 1 of the frame we have dDEedj 
and dD'Eted forms a polygon in the diagram. 

Failing Cases. — ^Whenever we find on arriving at any joint, 



22 



EECIPKOCAL DIAGRAMS. 



more than two unknown forces, this method apparently fails, 
since an indefinite number of polygons may then be drawn with 
corresponding parallel sides. A careful inspection of the truss, 
in connection with a sketch of the general outline of the diagram, 
will often reveal some condition fixing the relations of the lines 
in the diagram. 

This is the case in Fig. 77, A. M. " Arriving at joint 5, we find 
the forces parallel to" 5 4, 5 7, and 5 1 unknown ; but by sketching 
a diagram for this joint, we see that the lines representing the 
stresses on 4 7 and 1 7 must overlap or be portions of the same 







^^/- 




c 


^^^^^^^^ 


r"^ 


3?- 


fi^^^^^ 




y^ ^^^^ 




F 




Fig. 3. 



Diagram of stresses for half the truss reioresented hy Fig. 77, A, M. 
Letter that figure in a manner illustrated hy Fig. 1. 

line, in order to read correctly for the joints 4 and 7 ; and since 
5 7 and 7 1 make equal angles with 1 3, the lines on the diagram 
representing their stresses must intersect midway between lines 
drawn through e and/, parallel to 1 3. Hence, as CD is the line 
to be drawn, we may draw the line from C, parallel to 5 4, to a 
point D, haK way between the parallel lines above mentioned. 

There will be no difficulty with the remaining part- of the 
problem. 

Or otherwise : Since the stresses on 5 7 and 5 6 are seen to be 
equal, and make equal angles with the rafter, we may draw an 
auxiliary line from B parallel to 5 4, till it meets a line through 

Note.— For an extended discussion and application of these principles, the student 
is referred to " Economics of Construction," by R. H. Bow, C. E. 



ARTICLES 160-169. 23 

e parallel to the rafter, which gives E. From E and C draw 
lines parallel to 5 7 and 5 4, which will intersect in D. It will be 
seen that this makes ED = BC, or the stress upon 5 7 equal to the 
stress upon 5 6. 

Problems. — Draw diagrams for figures 79 and 82, A. M. 

Article 160. 

Example 1. The Tg^ of the first, and T^g of the last equations 
at the end of this example, should be omitted. 

Article 167. 

The funicular polygon is defined in Article 152 instead of 
Article 150. 

Article 169. 

Equation 8. — Combine the equation x^ -\- x,, z=z a, with the 
equation obtained from the proportion Vi'. ViW A '' A- 

Equation 9. See equation 5 ; m = -i- = — ; x-^ = 4. my ^^ 

/p 2 I ^2 

xl= i my^ .'. x,' ^ xl = 4=m (y, + y,), .-. m = / \ - 

^ yVi ~r 2/2) 
The remaining part of the equation can be obtained by sub- 
stituting the values of x^ and x^ in equation 8. 

Equation 16. s ■= / ds = I a/ d x^ -\- d y^ z= I d y 

From the equation of the curve, y? ^=1 ^m y\ -— — =1/ _ 

dy x ^ y ' 

.',s = Jdy y— =2 J dz Vm-]rz' in which z^ = y. 

Then by Integral C, we have 

5 = 2 1 1 A/^^r+^+ 1: log, (z + v^^ThT^) + c [ . 

= A/y Vm + y 4- m log, ( ^/y + Vm -f y ) + C. 
When y z= 0, s = 6. .-. = m log, Vm -f C or C=— m log^Vm, 



24 ARTICLES 169-172. 



.'. s = Vy {m + y) -^ m (log, (y'y + V^n^rV) - log, ^m). 






2 

y 2/2 _|_ ^ = ?|/_^ _[- 1 = m tan ^ . sec i See equation 6. 

.-. 5 = m •! tan z . sec z . + logg (tan i + sec ^) [• . 

■ To obtain equation 17 develop i \- \\ .by the binomial 

formula, multiply by dy and integrate two terms of tbe series. 

Abticle 171. 

The last of equations 1. Place the letter G at the intersec- 
tion of AB and BF prolonged. 



W 



Then DE = J EF = ^ EG ?__/since EP : FG ; : e/ 



' 2 



2 
(since E G = J B C, for A E = ^ A C). 

At the top of page 171. 

Place the letter A in its proper position in Fig. 86. y^ is the 
projection of the line DE upon a hne drawn perpendicular to 
AC. 

Article 172. 

Equation 11. In Fig. 87, A. M., draw hues through B and 
X perpendicular to the arrow line through P. It will be seen 

that ^ . cos e = - . cos j .-. ^ = arc . cos I -- cosyl • 



ARTICLES 172-174. 25 

Equation 12. Equation 16 of Art. 169, gives the length of a 
parabola, from its vertex to a point whose tangent makes the 
angle i with the axis of x, tangent at the vertex. Equation 1 2 
gives the length of that portion of the curve included between 
two points where the inclinations are i and j. This equation is 
obtained directly from equation 16 Art. 169, by taking that in- 
tegral between the limits i and / 

Equation 13. — In equation 17, Art. 169, substitute a: -f- 2/ sin 
/ for X, and y cos j for y. These values are obtained by an in- 
spection of Fig. 87, A. M. 

Article 174. 

„ . , ^ 6? u u 2 du (Pu 2 udu , /du\ 



"m- 



dlL^ _ U^ 1 _ ^ '^^ ^^ 



Let V^^' -\- a' c = z -\- u 
then u" -^^ a" c = z" -{-2 zu -{- u^ ov a^ c = z" -{- 2 zu. (A) 

Differentiate and divide by 2. 

z -X- u 

0^= z d z 4- z d u 4- 10 d z OY d u = — ■ d z 

' ' z 

.'. d X — —a .'. X = — a logg z -\- a log^ Cj. In which 

a logg Cj is the constant of integration. 

.•. — = - log, z -{- log, q = log, -j- • 

then e = — . •. 2 = = c, e 



From equation A above, we have u = 



a' c — z' 



2 z 



- 1x 

cC- c — c? e " «2 c 



— a: 



2 Cj e 



2 Ci 2 



Equation (c). — From Equation 1 we have —. — = y. Hence, 

differentiate equation (5), and divide by c? x ; the result will be 
equation (c). 



26 ARTICLE 175. 



Equation 6 is derived from (4), by multiplying by 2 e", 
which reduces it to the form 



2/o yo 



2/0 



i/-^-l ; ^ = log, /^+ i/-^--l\ 



■■""-•(,7 + ^lH)' 



Article 175. 

Equation 4. — By squaring both members of equation 3, and 

dx d X ^^ 

solving with respect to —. — we find —. — = / — ., ^ • The 

numerator of the second member is ?n, and not m^. 

Equation 5. — Integrate equation 4 by means of integral B. 
Equation 8. — Divide both members of Equation 5 by m^ and 

- ^ a/ ^ 

write the resulting equation in the form e "' — — -j-T 1-|-— • 

s 
Substitute the value of — from Equation 6, for the first term 

1 / ^ -^\ 

of the last member, and transpose. The result is-le»»-|-e "»■ 

= i/ 1 H This value in the first part of the first value of 

' m- 

y (Equation 8), gives y = a^s^ -\- m"^ — m. 

Equations 11. — Combine x^ -\- x^ = h, and x^ — x^ = k. 

Equations 12. — Putting these values of x in Equations 6 and 
8, the following will be found to be the correct results : 

k 



7 '^^ 

J = s, — s. = — - 



'^ = Vx - y2- 






Note.— The reference in the last line of page 178, should be to Article 174, instead 
of 173 



ARTICLE 175. 



27 



Properties of the Catenary, 

I. The radius of curvature may be found by making proper 
substitutions in the formula 



(■+»)' 



d X 



or as follows : 



d s^ 1 

in which ds = {dx^ -\- d y'^y • 



d y 



d X d^ y 
From Equations 1 and 3, ^ ^ 
ds' 



d'y 
d X 



,. dx. d s 
d'y= :p 



= d. 



: .d X . d s =: m i -- — | = ' 
\ d X / 




Fig. 4. 

II. From the Equation 
2/2 = 5^ -j- m^, (obtained from 
8 by transferring the origin 
of coordinates to a distance 
m below the vertex,) it will 
be seen that, if a right tri- 
angle be constructed, having 
y for its hypotenuse, and 7n for the side adjacent to the foot of 
the ordinate, the remaining side will be s, or the length of the 
curve from the vertex to the point having y as its ordinate. 

d y s 
From Equations 1 and 2, we have tan ^ = / = 

a X m 

Hence, if the lines are drawn as B D and D A in Fig. 4, tan i 

; hence, the line A D is tan- 
n m 

gent to the curve at A. 

If we draw A C perpendicular to A D, we have from the 

similar triangles A B D and A^ C, m \ y \\ y \ A C . •. A C 

^2 i y V 

= = m I — I = m . sec- A B D = m . sec^ i = p. See 

m \ jn f ^ 

Equation 17. 



= tan < ABD = ^- = 



28 



ARTICLE 83. 





Y 


'^4^--^:/^ 




F 


-..^^^^^^^^ 


yyC 






L--'>/ 


y/^^^\ 




§ 




Aii r^ 


Vb' X 





f-fi/ 


/ 




E 


^ 


Fig. 6. 





TV. In any 

position of the 
rolling p a r - 
abola, a line 
drawn from 
the focus to 
the point of 
contact, will 
be normal to 
the curve de- 
scribed by the 
focus, since it 
is the instan- 
taneous radius 
vector of the 
describing 
point. (See A. M. at the bottom of page 401.) 

The following shows that this line is equal to m . sec- * : 
Let B, in Fig. 6, be any point in a parabola whose focal dis- 
tance is m, F its focus, and B E a tangent at the point B, making 
the angle (3 with the axis X. The angle E == < B (from 
Analytical Geometry), = 90^ — (3, from figure. 

When the parabola has rolled till B comes into X at B\ the 
line B E will coincide with X, the line F B will be at F^ B^ and 
make an angle = 90^ — /3 with X, since a line drawn from B 
to the focus F will retain its position relative to the tangent, 
and a line drawn through F' ± F^ B' will be tangent to the curve 
described by the focus, since, as before stated, F^ B^ is normal 
to that curve. 
Then /3 = ^. 
From the polar equation of the parabola, we have p = 

2 m 
=^== when e = < B F E. (See Church's Analytical 
1 + cos ^ ^ 

Geometry, pp. 128 and 123.) 

e = 180° - 2 (90° - /3) = 2 /3 = 

cos d = cos 2^=1 — 2 sin^ i 

2 m m 



: 2 i 



' P 2 (1 - sin'^ i) 
Therefore as F^ B^ = ^ 



: F B = F B'. 



. sec^ 2, and is normal to the curve 
described, it follows by the first and second geometrical prop- 
erties that this curve is a catenary. 



ARTICLE 181. 29 

Article 181. 
The Geo7netncal Construction preceding Equations 5, 
At this point we need to show : 
1st. That 2 6^1 = A + B. 
2d. That ap and aq are equal to A and B. 
3d. That by construction, the triangle q^ O'q has a right angle 

at 0'. 
4th. That q p is perpendicular to 0'«, and tangent to the ellipse. 
5th. That the angle p 0'« is that made by a perpendicular to 
the tangent qp, with the major axis of the ellipse. 

1st. 

Complete a parallelogram upon O'a and O'B', and call the 
vertex opposite 0', E. Then by construction, the point E is in 
the prolongation of O'm, and is a diagonal of this parallelogram. 

From Trigonometry, O^E' = (2 O'mf = O'W + OV -[- 2 
O'B' . O'a . cos B'O'a. 

By an examination of Fig. 89, A. M., it will be seen that 
/ B'O'a = / j. Since O'A' and O'B' are conjugate diameters 
of the ellipse, and O'a = O'A' by construction, let O'B' = A' 
and O'a = B'. 

Then we have from the preceding equation, (2 O'w)'^ = A'^ 
+ B'2 + 2 A'B' cosy. 
cosy = cos . (90° — A'O'B') = sin of the angle between the 

A B 
conjugate diameters A' and B' = - • and A'^ -J- B'^ = A*^ 

A B 

-|- Bl See equations 2 and 3 of Art. 157, Church's Analytical 
Geometry. 

.-. (2 O'my = A^ + B2 + 2 AB or 2 O'm = A 4- B I . 

2d. 
In a similar manner we obtain B'a = 2 ma = A — B, . •. ma 
A- B 



2 

o- r.. A. + B 

bmce mp z=z mq =. O m =z ! , 

A+B , A-B . 

ap = mp -{. ma = i + — = A, 

A+B A-B ^ 

aq r= m^q — ma = = B, 



30 AKTICLE 181. 

3d. 

From Geometry 2 m/ _^ 2 OW = V + 0^' = (2 0' m) 
= (A -H B)^ = J<i\ 

. -. as the square of pq is equal to the sum of the squares of O'^- 
and 0^]), the triangle has a right angle at 0'. 

4th. 

In the equation on page 187, A. M., c = secj = as 

will be seen by an inspection of Fig. 89, A. M. By construc- 
tion, — = = — = cos / = cos B'O'a. This last value 

' 0' B' c r c -^ 

is obtained from the figure, where it will be seen that/ = B'O'f^r. 

Then as cos B'O'a ri= -, it follows that the angle formed 

B 

by joining B'a, is 90° at a, also that as the angle at a is 90°, the 
line qj) is parallel to A'O', and therefore tangent to the ellipse 
at B', the extremity of the diameter conjugate to 0' A'. 

5th. 

Call the angle p O'a, W. Draw perpendiculars from a to 
O'p and 0'^, and call their intersections with these lines, E and F. 

Then aE = ap sin /_p = A sin (90 — B') = A cos B^ 
a¥ =z aq &m /_q z:^ B sin B'. 
OV = aW 4- aW' = A' cos^ B' + B^ sin^ B'. 

By comparing this equation with equation 13, Art. 95, A M., 
and the notes upon that equation, it will be seen that B' =r /jo 
0'(x = the angle that a normal to a tangent makes with the 
major axis of the ellipse. Hence O'p is the direction of the 
major axis, and O'q at right angles with it, the direction of the 
minor axis. 

Equations 6 are obtained from the equation of the length of 
any diameter of an ellipse, in terms of the axes and the angle 
made by the diameter with one of the axes. 

Let k be the angle made with the major axis, and k^ the angle 
made with the minor axes. 



Then c' r' =.\/ ^' ^' = \/~- ^^^ 



A'^sin^^H-B^cos'^^ ^ A' co&' k' -\- B^ sm" k' 
(Church's Analytical Geometry, page 169.) 



ARTICLE 182, 



31 



The last value in equation 6, which is the value of sin // ob- 
tained by solving the above equation, should be 

A 



V- 



;2 y.2 _ B2 



Article 182. 

Theorem I. — Let 5^ represent 
any flexible and inextensible cord, 
fixed at 5i and in the plane X Y, 
and acted upon by forces in that 
plane. 

Let p be the force per unit of 
length of the curve, and a and /3 
the angles that p makes with the 
axes OX and OY. 

Let T = tension at any point, 
making the angles a and h with 
OX, OY. Let To = tension at 0, 
making angles «„ and ^o with OX, 
and OY. Let s = any length of 
the curve measured from 0. Let 
Si = whole length of the curve 
measured from 0. 

Kesolving all the forces parallel 

to the axes I since cos a 



d s 



and cos b = --- 

a s , 



, we have 




, ds . cos a — To cos «o = 0- 



To cos L 



as «./ s 

T — ^ + f'p . ds . cos Q 
d s t/ 8 ' 

Differentiate equations 1 and 2. 

m 7 id x\ dx , ^ 

T • c? I — - I 4- — - . dT = — p . ds . cos a. 

\a sf d s 



yd s 



'4- 



d 



IT = — p . ds . cos 13. 



Multiply 3 by ^^, and 4 by -f^, and add. 
as ds 



(1) 

(2) 

(3) 
(4) 



: ~pC 






32 ARTICLES 182-185. 

Then (i T = — i? (cos a . cos x -f- cos 5 . cos (3) ds = — p . ds . 
cos 0. (5) 

In whicli cos . = tlie cosine of the angle between the direc- 
tion of the force p, and a tangent to the curve. For in general, 
cos d = cos (a — «) = cos a COS a -}- sin a sin a = cos a cos a 
-\- cos 13 . cos b. 

If the force p is normal to the curve, then d = 90^, cos = 0, 
and T will be constant, since its differential is 0. 

In this case, equations 3 and 4 reduce to 

T ,dt^\ = - p. ds. cos a] T .d^^\= - p. ds. COS fi. 

Square and add the above equations. 
T^ 1"^^/. ^y + i f^l^Yl =P' ds' (cos^ a + cos^ /3) = p'ds\ 

T =^.-^=t!L_ = i)p. (6) 

y(d^'-\-(d^yY 

In which p is the radius of curvature. (See Church's Cal- 
culus, Art. 102.) 

Equations 2. — See Church's Calculus, Art. 106. 

Article 183. 

/^i x^ x^ 

X dx = w — — - — —t 
a-O 2 

/^l x"^ X^ 
xdx = w — ^—T 

Article 185. 
The equation following 12. When i = 0, the fraction 

— r^^-. is indeterminate. Its value is found by 

2 sm^ t 



ARTICLES 187-197. 33 

taking the third differential co-eflBcient of the numerator and 
denominator, and introducing the value i = 0. The result is 

— - • Hence for ^ = 0, ^^ = w r im — o I ' 

Equation 13 is found in a form more available for practical 
use, on page 423 of Kankine's Civil Engineering. 

Equation 16. — The cos/ of this equation should be sin/. 

Article 187. 
Equation 4. — In Equation 2 we have Ho := max. value of 

P^ . , " max. value of =r " • Differentiating both numer- 

dy 

ator and denominator with respect to y to find the value of this 

d. P, 

d y 
vanishing fraction, we have H^ = — (when y =: 0). But 

dy.dy 

dV -^ 

for 2/ = 0, "" = Pq. Hence, Hq = d^ x (for y = 0). 

d y 7 Q 



Equation 2. P sin : 



Article 191. 
P 



cosec V 1 + cot^ 



^ ^ tan^^ '^ ^/^ 

Article 197. 

Equations 5 and 6. — The last term of the denominator of the 
last fraction of each equation should be cos^ 0. 

T-r . ^ -r 1 . 2 w^ cos B — Px — p. 
Equation 9. — in the equation cos 2 ;// — • , 

substitute the values of ^9^, ^„ and p^^, from Equations 1, 7, and 8, 

, cos^ Q 1 

and reduce to the form cos 2 ;// — ; 4- 

^ sm ' 



34 . AETICLES 197-199. 

(1 — cos^ e\ 
sin <j> f 



cos d VCOS"^ 6 — cos''* (h ^ ^ cos^ 

^ But 



sm (j) sm(j) \ sin 

sin d 



— sin - . 

sm 



sm^e 



Vcos' — cos' (f) _ ./ (I — cos"' <^) — (I — cos' 6) _ J _ 

sin ~ ^ sin^ "~ ^ sm^ 

,/ sin^0 . sin 
. •. cos 2 ;L = cos y 1 —. sm -^ 

^ ^ sm' sin 

. -1 sin . . . -1 sin 

= cos . cos sm — sm . sm, sm 

sm sm d) 



(-1 sin 0\ 1 / . 

+ sm ^ I .-. J/ =r - I 4- sm 
' sin (/)/ ^ 2 \ 



1 sin 0\ 



L (^/ ^ 2 \ ' sm (j)/ 

By changing the construction of Fig. 57, A. M., as directed 
in Case 2 of Art. 112, to adapt it to this case, it will be evident 

A 
that 2 X n^ = 2 \p, SiS measured in that figure, is > 180° .-. -^z is 

> 90°. Therefore sin -. must be greater than 90°, in 

^ sm ° 

order that Equation 9 may be true. 

The reference near the bottom of page 216, should be to 
Problem IV, &c. 

Article 198. 

Problem. — Find the pressure against a vertical wall 12 feet 
high, sustaining a bank of earth which slopes backward from 
the top of the wall at an angle of 30°, the angle of repose of 
the earth being 45°, and its weight 100 lbs. per cubic foot. 

/3 V3~v 

Ans. 3600 I— = J = 1671 lbs. nearly. (Length of wall 

== 1 foot.) 

Article 199. 

Problem. — A prismatic column of solid masonry 80 feet high, 
and weighing 120 pounds per cubic foot, is to be built upon a 
foundation of earth whose angle of repose is 30°. What is 
the least depth below the surface of the ground at which the 
foundation course should rest, if the surface of the ground is 
horizontal, and the earth weighs 100 lbs. per cubic foot ? 

Ans. 104 ft. 



ARTICLES 199-215. 35 

Equation 5. — Draw a trapezoid whose base h is horizontal, and 
whose parallel sides are vertical and equal to p' and w a;, respect- 
ively. The ordinates of tliis trapezoid will represent intensity 
of pressure at any point, and its area, the total pressure upon 
the surface. Divide the trapezoid into a triangle and a rectan- 
gle by a line through one extremity of the shorter vertical, par- 
allel to the base. The distance c & is obtained by equating the 
moment of the whole trapezoid about the center of J, to the 
sum of the moments of the above mentioned rectangle and tri- 

angle. The moment of the rectangle = .-. 5 . . b c 

p' — XD X h 

- . . 



~ 2 • • 6 

v' — "^ ^ 

c = ^ , . — ^ — • • In this equation introduce the value of p' 

6 (p -\- lu x) ^ ^ 

as given in Equation 2 of this article, and the resulting equation 
will be the last value of c given in Equation 5. 

Article 202. 

Equations 10 and 11. — From what precedes these equations, 
it would seem that they should be obtained by substituting 

5 11 

■ — r- and — r- for - in Equations 5 and 7 respectively. 

S — |— i S — |— i A 

The following are the results obtained by such a substitution : 

s 1 

—-, — ; — TT- w t (x^ -\- h x) and ——- lu b (x^ 4- h x). 

2 (s -[- 1) "^ ' ^ 5+1 

Article 214. 

To obtain Equations 4 and 5 from Equations 1 and 2, divide 
the numerators and denominators of the second members of 
those equations by x, and then make x — infinity in the result- 
ing equations. 

Article 215. 

Equation 7. — Conceive of a circular chimney whose external 
and internal diameters are t and ^ — 2 B respectively. The 
outer circumference is tt t, which, multiplied by b, and placed 
equal to the sectional area of the masonry, gives 

.Je^^-^(*-2Er... i = B(l-l). 



36 ARTICLES 217-260. 

Article 217. 

Note. — The arm of the couple will be easily obtained by draw- 
ing from D and F, Fig. 101, A. M., lines perpendicular to H P; 
from F a line parallel to H P, and a horizontal hne through D. 

The first of these Hnes is — cos 6. The remaining term [q -|- i) 

t . sin (0 -f- i) is easily obtained. 

Article 225. 

" Find the center of gravity of the load between the joint of 
rupture C, and the crown A, and draw through that center of 
gravity a vertical line." See Fig. 107, A. M. "Then if it be 
possible, from one point in that vertical line, to draw a pair of 
lines, one parallel to a tangent to the soffit at the joint of rup- 
ture, and the other parallel to a tangent to the soffit at the 
crown, so that the former of these lines shall cut the joint of 
rupture, and the latter the keystone, in a pair of points which 
are both within the middle third of the arch ring, the stability 
of the arch will be secure ; and if the first point be the point of 
rupture, the second will be the center of resistance at the crown 
of the arch, and the crown of the true line of pressures." 
Rankine's Civil Engineering, page 442. Let the student make 
the above calculation and construction for an assumed circular 
arch. 

Article 234. 

Example 1. — The value of P^, is obtained by remembering 
that the thickness is small compared with the radius of curva- 
ture, and that the surface of a zone is equal to the circumference 
of a great circle, multiphed by the altitude of the zone. 

Article 235. 

Equation 5. — This equation will be found in Chauvenenet's 
Trigonometry, Formula 53, Page 163. 

Article 249. 
See reference made in Article 263, to this Art. 

Article 260. 

After CD in equations 1, read a x. At the end of next to the 
last line on page 282, read, principal elementary strains. 



AETICLE 271. 37 

Article 271. 
Equation 1. — See Art. 179, Theorem. 

Article 273. 

Equation 9. — In Equation 6, let jOq r=/ divide both numerator 
and denominator of the last value of p^ by r^, and solve the resul- 

13 

ting equation with reference to 

r 

Figure 119. — The general equation of Hyperbolas of Higher 
Orders, is y'^ x^ = a. 

X y =1 a is the equation of a h5rperbola of the first order, 
referred to its asymptotes, xy^ = ais the equation of a hyperbola 
of the second order. 

. •. xy^ =: a = ic' 2/'^. (A) 

Take R as the axis of Y, and a vertical through 0, as the 
axis of X. 

Then we have 

X '. x' \\ y'"^ : 2/^ from equation (A) 

r"A : R"B: [R^ : r" 

: : 0~R2 . 072 

By comparing these proportions with equation A, we have 

^A. X ^' = RB X R' = a. (B) 

The area r ABR ^ / / dx . dy = I xdy= I —^dy 

= ''('^-)'=''(7-i)- (C) 

= r-S- = -^ 1 .AXr-RBxR. (D) 

By comparing equation C with the last equation in case 2, it 
will be seen that this area represents case 2, and that a == 

tA X r\ (E) 

From the proportion C A : D B ; ; ^0 : g'l , we have C A ^^i 

= 065-^. By introducing the values of q^ and q^ , as found in 
the equations at the bottom of page 292, A. M., the equation 



l See A. M. 



38 ASTICLE 277. 

becomes (~ - m\ C~A = i^ ~ m^ DB = (RB - m) C A 
= (r"A - m) DB 

RBxCA-^xDB = m (CA - DB) = m (TA-RB) 
RB {TA - ^) - (r A . RB - TC) = m (r"A - RB) 
r~C (r~A - RB) = m (TA - RB) 
.'. m = r C. 
Hence it will be seen tbat r C represents m of case 1, in the 
solution of that case. 

Since r C is m, and r A X r^ = a, 

rA X ^ -TS 



We have from equation 4, ^ = 



/2 /2 



But r A : cc' : : /^ : r^ ... ^^ = a/. 

.-. 2 = cc' — r C r= the segment of the ordinate correspond- 
ing to 2/ = /, between C D and the curve A B. 
Again, p=z—^-\-m = x'-\-rG = x^-\-r'Ei. 

= the entire ordinate from E F to A B. 
From C, we have 

^, =^+m = lA_><Z!.+ r-C = rA + .-^= AS. 

Article 277. 
Differentiate Equation ] . 

w s dx=fds ; -— dx = — . Then — — = log^ S + C. 

/ 5 / 

W 
When x = 0,S = ; see equation (1) A. M. 

W W 

Then = log, -^+ C or C = - ^og, — . 

W X ^ ci 1 W , /S 

.-. _ = log, S - log, — = log. 4^ . 



• AETiCLES 295-298. 39 

Aeticlk 295. 



Talle. — Read m' = \, instead of - • 
A y 



Article 298. 



Equation 1. Let Gj , G2 , Gg, and G be the centers of gravity 
of Ai , A2 , A3 , and A respectively. 

Take moments about a horizontal line passing midway between 
the top of the upper flange, and the bottom of the lower flange. 

Ai = A, fh±]}l±Jh _ i?) _ A, fll±J^±h - ii\ 

= A, (^-1+^2) _ A. (^) - A, (^^ . 
..2/._--x_- . ^^ (1) 

Equatioru 2. 
' 2 2V 2 ^ 



^ A, {\ 4- ^. + 2 A3) + A3 {\ + ^3) 
2 A 

Q-Q. __ h f K + K K - _ Ai + A3 _ - 
'22 2 

_ Ai (A, + A, + 2 A3) + A3 (A , + A3) 
2 A 

G7G = ^±4±^-(A,+i!)+^=^^^- + ^ 

_ K {K + ^^3) - A^ (A^ 4- A3) 
2 A 

The moment of inertia of the section, about its neutral axis 
will be 

I = ^■^'■' + ^^y + ^^^° + A, X g7G^ + a, X ^G^+A3X GTG'. 

See tlie end of Art. 95, A. M. 



(A) 



(B) 



(C) 



40 AKTICLE 298. 

... I == ^2jh±^^K±^]^ + 4X[^^ ^^ (^^ + ^. + 2 Asy 

+ 2 Ai A^ A3 (Ai + ^2 + 2 A3) (^ 4- ^3) 
-h Ai As'^ (Ai 4- hy, + A, V (/^i + ^2 + 2 ^3)^ + 2 Ai A, A3 

(Ai 4-^2 + 2 A3) (^2 + ^3) 
-1- A, Al (A, + h,)\ -L- A3 Al (A, + h,y - 2 Ai A, A3 (A, + A3) 

(A, + ^3) + A3 A,^ (A, + A3)^]. (D) 

But 2 Ai A2 A3 (Ai + A2 + 2 A3) (Ai + A3) = 2 A^ A2 A3 (A^ + Ag)^ 

+ 2 Ai A2 A3 (A2 4- A3) (Ai + A3). 
2AiA2A3(Ai-f A2+2A3) (A2 + A3) = 2 AiA2A3(A2+ A3) 

+ 2 Ai A2 A3 (h, + A3) (Ai + A3). 
.-. 2 Ai A2 A3 [(Ai + A^ 4- 2 A3) ((Ai + h,) + (A^ + A3)) - Qi, + A3) 

= 2 Ai A^ A3 [(A, H- A3)^ -f (A, 4- A3) (A2 i- A3) + (A^ + A3)^. 
= A, A, A3 [(Ai -f- hy 4- 2 (Ai + A3) (Ai 4- A3) + (A^ + h,y] 

+ Ai A2 A3 (Aj .+ A3)2 4- Ai A2 A3 (h, 4- A3)^ 
= A, A, A3 (Ai 4- ^2 4 2 h,y 4- Ai A2 A3 (h, 4- A3)-^ 4-Ai A^ A3 

(K 4- /^s)^ 
This reduces equation D to the form 



1 = 



AiA^-4-A,A./4-A3A3^ 



12 4-^^[(^^i + ^^H-2A3)^(A,A^, 4- 

A, A,' 4- Ai A, A3) 4- (^1 4- A,)^ (A, A3^ 4- A3 A,^ 

. 4- Ai A2 A3) 4- {K + ^3)^ (A2 A3^ 4- A3 A/^ + A, 

A2A3)]. (E) 

AiA2^4-A2A^ 4AiA2A3 = (Ai4-A2 4-A3)AiA2 = AA,A2. 

Aj A32 4- A3 A,^ 4- Ai A2 A3 = A A, A3. 

A2 A32 + A3 A^^ 4- Ai As A3 = A A, A3. 

These values reduce equation E to the form 

A7,2_1_A7, 2_i_A7. 2 1 

j._ A,A, +A,A, 4-A3/ Z3_ _^ _L_[A, A, (A, 4- /^2 4- 2 A3)' 
12 4 A 

4- Ai A3 (A, 4- hy 4- A2 A3 (A^ 4- ^3)^. (2) 

Equation 3.— See Art. 294, A. M. 

Equation 4. — This equation can be obtained by considering Aj, 
A2 and A3 as being so small that they may be left out of consid- 
eration. We should then have A = A^ 4- Aj and A' = A3 ^ A. 
These values in equations 1 and 2, give 



. ARTICLE 300. 41 

2'' = T 2A =T\ A ; A'-(^) 

1 7/2 A A 

I = 4^ (A, A, (2 hj) = •2- ' • (6) 

Introduce these values in equation 3 

A.J /'. A, A, 

Then M, = m WZ = Al = ^7^;^^ =/, A' A^. (4) 

Article 300. 

The last of Equations 2. — To obtain the last value of 

1 M / M 

-, multiply -^ J by z= — , and divide by its equivalent „ ^ • 

IX. The value of n". By making the proper substitutions in 
Equation 5, we have 

n" c^ = I I d x^ =z c I lop;. d x. 

t/ot/oC — X U a ^^ c — X 

Integrate by parts . I udv =z uv — I v d u :=z the general 

formula. 

Let loa:, =: u and d x = d v. 

^^ c — X 

rr^ #1 " -I " I X d X 

Then c 



'■ I lOff- =z C X lOff . c I - 

t/o^^c — X ^ c — X t/o< 



DC — X 

Let c — X z=^ z. Then d x z=. — d z\ x =: c — z 

/" X dx f*" d z ic — z\ , ) ° 
■ — — I ^ =z — G log, Z -\- z\ 

= I - c log, {0- x)-\-{g -x)\^^ 

••• fy^^^T^ dx = \cx .\og,-^ ^cnog,{c - X) 

-c{c-x))^^ 

:=z c \x loge c — a: log, (c — a:) -f c log, (c — x) ~ (c — x)]- 



42 ARTICLES 307-312. 

= C \ X loge C -f (C — a?) loge (c — x) — (c — x) ^ =:C 



.-. n" c' = c' or n" = 1. 

Vd" — x' 



X. Value of n" & = I , = c / c^ a; 

t/ e/ V/-2 V2 e/ 



. -1 a; 

sm 

c 

Then — = sin y ; d x z=z c cos y d y 

/c -1 a; /'" 

c/ ic sin — =: c^ / y cos y d y =i (hj parts) c^ 

4 

•! 3/ sin 3/ — j ^m y dy I = c'^ -! ?/ sm 3/ + cos y t =: c^ 

\ —1 QC- —1 iT* —1 1* f *' 

i sin — sin . sin \- cos .sin — r = 

( c c c ) 



■•-"=i-i- 



Article 307. 



Remark. — Near the bottom of page 334, read, m" is never less 
than J. 

nfh h} 
The last member of Equation 3 should be —j. ; • 

Equation 13 should be M — Mj = — ^—- — M^ = 

w (c^ — x^') S w c^ ' 3 lu c^ /I 4 x' 



(1 4 x2\ 

3 ~ T^}' 



Article 309. 



Equation 3. — Read / ^ y z d y for I y x d y. 

t/ t/ 

Article 312. 
Equation 7. — Read — for the co-efficient of the second mem- 



AKTICLES 314-318. 43 

v" 18 2 

ber. Also for — ^ — (below 7), we should have -^~ — -p— - 

nearly. 

Article 314. 

In the first place, suppose the external load "W^ is made up, 
in part, by the weight of the required beam. Then, having cal- 
culated the breadth, and found the weight B^ of the beam, it 
follows that, if W^ represents the whole load, W^ — B^ will 

represent the load exclusive of the beam, or ^^^r^^ r—- = the 

ratio of the gross load to the load exclusive of the beam. 

Equation 1. — The student needs to bear in mind that h^ is the 
breadth to bear the gross load, minus the weight of the beam, 
and W^ is the external load. Since h is a fixed value in this 
case, we have from Mq == mW I = nfhh^ (see page 316), and 
above ratio, ^' : W — B' y, h : h' 

; ¥W 

Equation 2. — The depths being constant, the weights of the 
beams are proportional to their breadths. 

...B.:B::..:J::..:^^g,::i:^^...B=. 

B^ W^ 
W^ - B^ 

Equation 3. — The true gross load will be 

W^ -4- B = W^ 4~ — — — ■ = /3^ 



Article 318. 
Equation 5. — Prom (4) 
d'x P 2dx.d^x P 

(47)^-^-^ + ^- When.= .,4|-=0...C = a-^ 



(2) 



i^Y 



44 ARTICLES 314-318. 



d 






dy ' -\ X y , n 

. ^ .'. sm == jL -f C. 



"When ic = 0, 3/ = .*. c = 0, and the above equation becomes 

. ~\ X y X . y . y 

sm — = — or — = sm — .-. ic = a sm — • • 
a c a c c 



STRENGTH AND DEFLECTION 



OF 



HORIZONTAL BEAMS. 



STRENGTH OF BEAMS. 



NOTATION. 

I = Length of beam. 

c = J the length of a beam supported at both ends, or the 

length of a beam fixed at one end, and free at the 

other. 
h = Breadth of beam. 
h = Depth of beam. 

y = Distance from neutral axis to outer fiber of beam. 
w = Load per unit of length of beam. » 

W =■- Total load. 

R = Force at right hand support. 
L = Force at left hand support. 
M = Moment of external forces, at any cross-section. 
Mq = Maximum value of M. 
F = Shearing force at any cross-section. 
F** = Maximum value of F. 

I = Moment of inertia of cross-section, about its neutral axis. 
E = Modulus of elasticity. 
/ = Intensity of stress at the outer fiber. 
2/1 = Maximum deflection of beam, 
m = A constant factor depending upon the method of loading 

and supporting. 
n = A constant factor depending upon the form of the beam. 
Jc = CoefiBcient of maximum deflection. 
= The Origin of a system of rectangular coordinates, in 

which the axis of the undeflected beam is taken as the 

axis of X. 
The following notes on the strength and deflection of beams, 
contain the substance of what is found in the A. M., from Arts. 
288 to 308, inclusive, and may be substituted for those articles. 
It is believed that, from the simple demonstration of the fun. 
damental formulae for the restricted case of horizontal beams 
with parallel vertical loads, the average student will proceed to 
their application more rapidly and with greater independence 
and security than by following the Applied Mechanics. These 
pages are therefore intended to be complete in themselves. 

Rankine's notation has been followed as far as practicable, 
and whenever introduced the letters stand for the same quanti- 
ties as in the A. M. 



STEENGTH OF BEAMS. 47 

All the notation will be found explained on page 46, or in 
connection with the subject where it occurs. 

In all problems relating to the strength of materials, two dis- 
tinct systems of forces must be considered. 

1st. The external forces, consisting of the applied loads, and 
the supporting forces. 

. 2d. The stresses, or forces acting amongst the fibers or par- 
ticles of the material. 

As these two systems of forces stand related to each other as 
cause and effect, it follows that, so long as equilibrium exists, the 
two systems must balance each other. 

If we conceive of a cross-section made at any point of a beam, 
we can imagine the beam broken by the particles sliding upon 
each other, along that section (shearing), or by the pulling apart 
or crushing of the fibers in a direction perpendicular to the sec- 
tion (cross breaking). Since a part of the fibers are extended 
and part compressed, these stresses, of opposite kinds, form a 
moment about the neutral axis of the section which tends to pre- 
vent the beam from bending, and which is equal to the moment 
of the external forces, or "bending moment," M. 

To find the value of F, take the algebraic sum of all the ex- 
ternal forces acting upon the portion of the beam included be- 
tween the section and one end, considering the upward or sup- 
porting forces as positive, and downward forces or loads upon 
the beam as negative. 

M is found by taking the moment of all the external forces 
actmg on the beam between the section and one end, about the 
neutral axis of that section. "We have found it convenient in 
most cases to use right hand rotation as negative, and left hand 
rotation as positive. 

The maximum values (FJ of F, and (M^^) of M, are generally 
required. The former is evidently some part of the whole load 
(W), and the latter some portion of the product of the load and 
length of the beam, and is expressed by the equation M^ = 
mWl. 

Beams of sufficient strength to resist the bending moment, 
have usually ample strength to resist the shearing force ; hence 
any further consideration of it here, will be omitted. (See Art. 
309 A. M., and Table II in the Appendix.) 

To determine the manner in which the beam resists the bend- 
ing moment M, notice that when the beam is slightly curved, the 
layers of fibers upon the convex side are extended, while those 
upon the concave' side are compressed, so that there will be one 
surface in which the fibers are neither extended or compressed. 



48 STRENGTH OF BEAMS. 

This is called tlie " neutral surface." It is assumed, (which is 
near enough to the truth for most practical cases,) that this sur- 
face passes through the center of gravity of the cross-sec- 
tion of the beam, and contains the neutral axis of that section. 

Within the limits of proof .stress, (see Art. 245 A. M.,) the co- 
efficients of extension and compression are exactly or nearly 
equal. From Hooke's law that the strains are proportional to 
the stresses producing them, we have zero as the stress at the 
neutral surface, and at any other point it is proportioned to the 
distance of that point from the neutral surface. 

Section 1. — Moment of Bending Stress. 

Let p = the intensity of stress at any distance y, from OZ the 
neutral axis of the section. 

Let/ = intensity of stress at outer fiber. 
d y .d z = an elementary area. 
p . dy . dz z=z stress on an elementary area. 
p . y . dy . dz =z moment of this stress about the neutral 
axis. 

/ / ^ p ' y * dy . dz . — \hQ total moment. 

Since the stress is zero at the neutral surface, and increases to 
/ at the outer edge, and is a uniformly varying stress, we have 

i^ :/. . 2/ : yoOYp = —y. 
This value introduced in the above integral gives, 
r^ r''^y'dy.dz= Ll (A), 

which is the moment of bending stress at any section. A few 
values of I are given in a table at the end of Art. 95, A. M. 

From these values it will be seen that I = a h h^, in which a 
is a constant, depending upon the form of section. 

Also, that y^ = dxh where c^ is a constant, depending upon 
the position of the neutral axis of the section. 

.'. ~ — = , , =n f h h\ where ^ is a constant and = -^• 

2/o ^^ ^ 

Hence the general formula for the moment of bending stress 
at any section is, 

-/^ oinfhli' (B). 



STRENGTH OF BEAMS. 
TABLE 1. 



49 



Values of 


n 


Rectangle h h, including square, 


1 
6 

TT 

32 

fA'- 

rA- 




Hollow Rectangle, h h on outside, and h^ h^ on 
inside, 


bh'f 


Circle and Ellipse, 


Hollow Ellipse, outside diameters h and h, in- 
side h and A, 


b U'f 


Hollow Circle, diameters h and h, 





For other forms of cross-section, calculate n from the formula 
1 



y b li' ' 



Section 2. — Transverse Strength of Beams of Uniform Section. 

Calculations upon the transverse strength of beams are based 
upon the principle of equality of moments. The maximum 
moment of the external forces must be equated to the moment 
of bending stress at the cross-section of the beam. 
Thus 

, Mo or mWl ) ( / I 

I Mo. of external forces ( =" } —^ = nfbTf ^ ,g. 



2/0 



Mo. of internal stress 



Problem I. — To find the safe load for a given beam. 

The modulus of rupture of ash being 12,000 pounds per 
square inch, find the safe load that may be uniformly distrib- 



50 STRENGTH OP BEAMS. 

uted over a rectangular ash beam whose cross-section h x ^ is 
3" X 12", and length / = 20/;., when supported at both ends. 

As the modulus of rupture is usually given in pounds per 
square inch (see Table IV in the appendix of A. M.), the cross- 
section and length of the beam should have their dimensions 
expressed in inches. 

Solution Z 3= 20 X 12 = 240" 

The whole load, W = wl =z 2ioc = 20 x 12 X '^^ = 240iy. 
The supporting forces R and L will each be tvc — I20iu. 

Let the origin be at the center of the beam, and take a sec- 
tion at a distance x to the right of 0. Then the moment at that 

section will be M = R (c — x) — lu (c — x) I | = 

The maximum of value of M will be obtained by making 
X = 0. 

Or Mo = Re — = wc^ ~ = -^ z= — W Z z= 

SOW. 

By Art. 247 A. M., it will be seen that Rankine considered 

10 as the proper factor of safety for timber, so that the outer 

fiber should not be under a stress greater than 12000 -^ 10 = 

1200 lbs. per square inch. 

h¥ f I 
.-. f z=. 1200. I = — — - Substitute in the equation M„ = 

and we have 

3x1728 
1200 X -V— 
30W = ^2— ^-^^ = 200 X 3 X U4. 

T 

9QQf| 

W ^ 2880 lbs., which is ^^— r- =144 lbs. per lineal foot 
of beam. Ans. 

Problem II. — Knoioing the form of cross-section of a given beam, 
and the load, to find its dimensions. 

If the modulus of rupture of cast iron beams is 40,000 lbs., 
find the cross-section of a rectangular beam 25 feet long, and 

Note.— The student mnst continually have in mind, that all weights or measures 
must be reduced to units of the same denomination. 
Note.— The subject of factors of safety is explained in Arts. 245-347, A. M. 



i STRENGTH OF BEAMS. 51 

whose breadth and depth shall have the ratio — = - , for a load 

of 200 lbs. per lineal foot, and a concentrated load P = 2200 
lbs. at the center, the beam being supported at both ends. 

Z = 25 X 12 = 300" . c = i- = ISO'' . W rrr (25 X 200) 

25 y 200 50 
-[- 2200 ri: 7200 lbs. to = _^ ^^ = -^' i Factor of safety 
Zo X 12 o 

=^6 yf = 4G000 ^ 6 =: -^—> n = -(See Table I) . Mo 



6^ 
3600. 



Let the origin 0, be at the center of the beam, and take a 
section at a distance x to the right of 0. Then we have 



=znfhh' = nfi~'Sh'; R = L = 

Let the origin 0, be at the 

section at a distance x to the ri 

w 
M = R (c — cc) — — (c — xy 

w 1 50 

... Mo = Re - -K- g' = 3600 X 150 - - -^ X 150^ = 150 
2 2 o 

(3600 — 1250) = 352500. 
... 352500 = - X ^^ X ^ X A' (FromMo=:r?i X/^T ^^'^ 



h = |/!M^^;5! = nearly 9.84". 



100 
b z=:--- = 3.28" nearly. 



Problem III. 



Find the strongest rectangular beam that can be obtained 
from a given cyhndrical beam. 

Let D be the diameter of the circular section of the given 
beam. We have from the equation of the circle, D^ = b'^ -\- h^ 
... ¥ = D^ — b^. Since for the transverse strength of beams 
Mo == nfb h\ we have Mo == nf(I)' b - b'). 

The value of b which makes Mo a maximum is found to be, 

_ ./2~ 
The corresponding value of his, h — D r o ' 



52 



DEFLEXION OP BEAMS. 



.1^ 



Section 3. — Equation at the Elastic Curve for Beams. 

Eesume the 
equation M := 

— Let s equal 

2/o 

the stress on a 

fiber at a unit's 

distance from 

the neutral axis. 

Then 5 : /" ; ; 

/ 
1 : 2/0 or— =8 

2/0 

Let X be the 
elongation of a 
fiber at a unit's 
distance from 
the neutral axis. 
Then d \ will be the elongation of a fiber whose original length 
was d X. 

Let jO be the radius of curvature of the neutral axis. Let E 
be the modulus of elasticity, that is, the force that will elongate 
a bar whose section is unity, to double its original length, pro- 
vided the elasticity of the material does not change. 

From Hooke's law, ''As the stresses, so the strains," we have 

5 : E " d \ : dx or s = —, — E. 

d X 

. /I ^ ^X 
" yo dx 

From the similar triangles in the figure above, we have 

d.\\dx 





\ 

\ 

\ 
\ 

\ 




4 


K. 





EL 



(2) 



But / 



1 


d\ 

•> ••• dx ^ 

/I 

2/0 
dxd^y 


1 

9 
EI 




(^ 


9 


dx') 

d'y 

dx' 



(3) 



p d x^ 



DEFLEXION OF BEAMS. 53 

d y 
Since —^ — is the tangent of the angle that a tangent line 

to the curve makes with the axis of .r, and in beams is exceed- 
ingly small when compared with unity, it may be neglected. 
Then we have 

... -^^-^ = E I ^ or M = E I -y4 (E) which 

is the equation sought. 

Problems. 

1. Find the maximum deflection of a beam loaded uniformly, 
and supported at the ends. 
Solution. (Origin at center.) 

^l—^^z=M.=wc{c — x)—iv{c — x)A^^. 

w 

d X 2 6 3 

■C, J d y IV C NO , w , ,, , w c* 

••• ^^5^ = -^(^-") +t(^-") +^ 

When X =z c,y = .:I) = — ^1^. 

6 ^ ^24^ ^ ^ 3 3 

At the point of maximum deflection — - = 0, or the curve 

d X 
is horizontal. In this case it is obviously at the center, where 

X z=z ^. 

— wc'^^ ?_ W l^ 

24 384 



.-. Ely, = t^ c'(- — -l l\ = _ 

"^' \6 24 3/ 



Note.— All the beams in this and the following section are supposed to be of uni- 
form cross-section, horizontal in position, and with vertically applied loads. 



54 DEFLEXION OF BEAMS. 

5 W P . 

•^' .384 EI 

2. Find the maximum deflection of a beam supported at the 

1 p /3 
ends, and loaded at the center with P. Ans. -— --—- 

48 hi i 

3. Find the maximum deflection of a beam supported at the 
ends, and loaded at the center with P, and uniformly with W. 

4. Find the maximum deflection of a beam fixed at one end, 

1 p ^3 
and loaded at the free end with P. Ans. — - — — . 

O ii/ i 

5. Find the maximum deflection of a beam fixed at one end, 
and loaded at the free end with P, and uniformly with iv. 

6. Find the values of M^^ and y^, for a beam fixed at its ends, 
and loaded uniformly with w, and at the center, v/ith P. Find 
also the point of contra-flexure. 

Let w l = 'W^ 

ElJ| = M = (l + ^c)(c-x)-^(c-..)^-^. (1) 

(from formula E.) where ^ is the moment required to hold 
either end of the beam horizontal. 

■r^, r. d y ^ ^ / P , 'M^ c \ „ w c^ T c^ IV c^ 

When.= 0,^=O...C = (-+_) e^-— =^+-- 

When=.=c,^=0.-.^c = Cor^ = ^ = ?-l + !i^ (2) 
d X ' c 4: 3 ' ' 

(-+-).+- + ^^ (3) 

(^c — xf + D. 

When X =z c, y = .'. D = 0, 
y is a. maximum when x := c. 

Note.— "Fixed," means fixed in direction. 



DEFLEXION OF BEAMS. 55 

(Y F w l\ 2P+ W,^3 
192 384 / ~ 384 

To find the value of M^, substitute the value of fx, from equa- 
tion 2, in 1 

(¥ + ^)- (^) 

Which gives a maximum value for M, where x == c, 

To find the point of contra-flexure, make --4^ = 0, in equation 

4, and we have 

(P V , , P c 2 c^ - 

P , _^ ,/P' , Pc , , Pc 2 c' P , 

+i/i:7ii;z. 



2W, 2— ^ 4W/^2W Tl2 



\2W, 2— '^ 12 W7 ' 



\2W^^2^2W/ 3 / 

= distance from end. 

7. Find the value of R; the point of Max. deflection (xY, the 
Max. bending moment; and the point of inflection for a beam 
fixed at L, supported at R, and loaded uniformly. 

R = f ^yZ = | W, 

ic^ = .58 I (nearly,) from L, 

Max. moment ^ W I, 

Point of inflection at ^ I from L. 

8. Find the pqint of contra-flexure, and values of R, ?/„ and 
Mo, for a beam loaded at the center with P, fixed at L, and sup- 
ported at R. 



56 DEFLEXION OF BEAMS. 

First consider that part of ttie curve included between ttie 
right hand support and the center. 

EI^ = R(2c_a.) = M (1). 

Bl4|- R (2 . - g;)^ + C (2). 

2 

■Ely=-^{2c — xyJrCx-\~-D (3). 

When X = 2 c, y = .: J) = — 2 G c. 

... E I y = -^ (2 c — o;)^ — C (2 c — x) (4). 

Now consider that part of the beam included between the 
center and the left hand support, 

^^ ji's=R(2<'-^')-I'('^-»:') (5). 

^ ^ "S" = - T (^ '-^'>'+ T (''-'''>' + ^ 

When x = 0,-^ = .-. F = 2 Re' ^^ = — (4R — P) 

{i-R--P) (6.) 

E 1 2/' = 1^ (2 — xj — -^ (c — xj + |- (4 R - P) 0^ 
+ G. 

When a;' = 0, 2/' = .-. G = g 1- —^ + y 

(P — 8 R.) 

.-. E l2,'= 4 (2 c-^'f - 4 (0-0,7 + -^(4R_P) 

+ 4 (P-8K) cy 

J dy dy' 

When jr = a;' = c , 2/ = 2/' and -^^ = -^-^. 

Under these conditions, we have from 2 and 6 

-il + C — 4i-4,.E-,) 



DEFLEXION OF BEAMS. 57 



.«. C = — (4 R — P), and equation 4 becomes 

El y=~ (2c-xY- ^{iR-T)(2c-^ x). . (8). 
"We also obtain from the above condition and 7 and 8, 

(P — 8 R) ; = (4 R - P) + J (P - 8 R) = 24 E - J P + 
P-8R.-.R= Ap (9). 

To find the point of maximum deflection, substitute the 
values of R and C in equation 2, and we have 

The maximum deflection is where — ^ — = 0, hence to find 

a X ' 

5 c^ 

the point, we have ^ = — ^ {2 c — xf -{- --■ 



.'. 5 {2 G — xf = 4.C 
(2 c - x) = ± 2c |/i 



5 



x = 2 c — 2 c yh = I /l —y- \ = distance from fixed 

end. The amount of maximum deflection will be found by sub- 
stituting this value of x in equation 8, which gives y^ 

(1-1(1-^1 ))) 

\96 '^ 125 32 '^ 5 / EI 

^ 5 \96 96/ EI 
4.8^ 5 El' 



58 DEFLEXION OF BEAMS. 

Substitute the value of R in equation 5. 
"^^"^^ Ti= ^P(2c-a=')-P(e-»/) (11). 

To find tlie point of contra-flexure, make -^-^ = 

Then ~V (2 c - a/) - V {c - x') = 

lOc — 5afz=l6c— 16 a/ 

6 6 

a/ z=z -- c = —I = point of contra-flexure. 

To find Mo from equation 1 1 we have 

P P 

M = -~ (10 c - 5 x' -{- Q c -\~ I6x')= -- (Uaf -6 c) 

Making x' = 0, we have 

3 8 
M — Pc— — PZ 

9. Find the same quantities for a beam fixed at one end, sup- 
ported at the other, and loaded uniformly, and at the center 
with P. 

10. Given a beam supported at the ends and center, with a 
fixed load over its entire length, and an additional load over 
one span, to find R, and the point of contra-flexure. 

Let w be the intensity of the fixed load, and w^ that of the 
additional load. 

The tendency of this load is to curve the opposite end of the 
beam upv/ards, so that instead of requiring a support it would 
have to be fastened down at that end. 

Let the left hand span be thus loaded, then R will act down- 
ward. 

Since the fixed load is distributed symmetrically with reference 
to the center, in calculating for R, it may be considered as all 
supported at the center. 

Taking the origin of coordinates at L, and the coordinates of 
the left hand x^ and y^, and those of the right hand span, x and 
y, we have the following: 
For the latter 

E 1^1^ = — R(Z — a:). 



DEFLEXION OF BEAMS. 59 

E 1 2/ = — I {l — xy + A a; + B. 
For the former 

^' 6 ~ 24 ~ ~ 

These equations contain six constants, A, B, C, D, R, and L. 
The six equations required for their determination, may be found 
from the following conditions: 

When x=-, y—^. When a;^=?, 2/^=0. 

" x — l, y — ^. " x^=0,y=0. 

^ "~ ^' ~" 2' 'Jx'~ d~xl 
I d'^y d' y^ 
' a dx^ d xf 
Forming the equations and eliminating we have R = — 



-- I — '— I = — the additional load. 
16\ 2 / 16 

L r= — I — '- — I =: W I. 

16\ 2 / 32 ' 



A=-^ 
24 



RZ 
~2r' 



f_y,_ 



To determine the point of contra-flexure, put — -^=0, remem- 

d xj 

bering that M must include the fixed load which is supported by 
the center pier. 



^^^^=-^^^ + (^ + ^^)v'=^ 



2 

7 x^ 

^, = -- — ^ — = distance from left end. 



60 



BENDING MOMENT AND DEFLEXION. 

TABLE 2, 



Beams of Uniform Section. 


Values of 


M„ = mWl 


Vi 


1. 

\ Loaded at the center with W. 
' Supported at the ends. 


i.WZ. 

4 


1 W P 
48 E I • 


2. 

j Loaded uniformly with w. 
i Supported at the ends. 


Iwi. 


5 WZ3 
384 E I 


( Loaded at free end with W. 
\ Fixed at one end, free at other. 


Wl. 


1 WP 
3* EI 


4. 
j Loaded uniformly with w. 
\ Fixed as in 3. 


1.W?. 


1 W F 
8 E I 


5. 

r Loaded with w, and at the 
extreme end with P. 
Fixed as in 3. (^v Z= WJ. 


|(2P + W,). 


i-p4^) 


6. 
j Loaded at center with W. 
i Fixed at the ends. 


>w. 


1 W P 


192 E I 


7. 
\ Loaded uniformly with w. 
, Fixed at the ends. 


h-^<- 


1 W P 
384 EI 


8. 
C Loaded uniformly with w, and 
} Sit the center with P. 


rt7^-- 


('>^V4-W,) P 


( Fixed at the ends. 


oo4 E I 


9. 
C Loaded uniformly with w. 
■ Fixed at one end and sup- 
' ported at the other. 


>■■ 


2.08 W P 
384 E I 

nearly. 


10. 
Fixed as in 9, loaded at cen- 
ter with W. 


tI-'- 


1 /SWP 
48r EI 



FORMULA FOR DESIGNING BEAMS. 61 

The preceding table gives the values of Max. bending moment, 
and Max. deflection for the most common and important cases 
of loading and supporting. In the column headed M^ — m W /, 
the co-efficients of W I which are the values of m for the different 
cases, are so arranged as to be seen at a glance and readily taken 

out for use. Also the co-efficients of -—-—in the -column headed 

E I 

y^ are the values of h in the formula ^/i = h. , or co-efficients 

E 1 

of Max. deflection. (Cases 5 and 8 are not included in the 
above remarks relating to the co-efficients). 

Section 4. — Equation for designing a heam whose deflection shall 
not exceed a given amount 

In the prece'ding problems we have seen that the value of y^ 

kW l^ 
may be expressed by the equation y^ = (1). 

E I 

Hencei = ^ (2). 

f I 

From formula 6 in section 2, we have M^or mW I = 

% 

/ * 
Substitute this value for I in equation (1). 
_ kW F f _ Jc f l^ 
^^ -WmWTi^ ~ m >^ ~B ^ ^- (^^• 

In beams symmetrical with respect to the neutral axis of the 

2h f l^ 
section, y. = ^h\ hence (3) becomes ?/, = — x 4r X -^' 
^ ' m ^ h 

yi 2k f I 
I m El h ^ ^ 

h 2k f I 
or 7 = — X % X — . (G.) 

« m E 2/i ^ ^ 

In applying this formula, k and m are to be taken from table 
(2), or calculated as heretofore explained. E and / are found 
in the Appendix to the Applied Mechanics, tables 1 and 4 re- 
spectively ; but the value of / there given, being the ultimate 
strength of the materials, it must he divided hy a proper factor of 
safety before being introduced into the formula. 



62 BEAMS OF UNIFORM STRENGTH. 

I 

— is the ratio of the length of the beam to the greatest 

deflexion allowable for the particular case in question, and is 
assumed from experience or from the nepessary limitations of 
the problem. 

Problem. 

Find the dimensions of the rectangular cross-section of a 
wrought iron beam 20 feet long, which shall deflect but 
T^Vn 0^ ^*s length, for a working load of 225 lbs. per lineal foot. 

5 
k = -^^ I See Table 2, 



384 



1 ^ 
n = g i See Table 1, 



E = 27000000. 
Working strength of wrought iron, 
/ = 9000. 



■■■i = zm' ^ - 12 X 20 = 240"; W =: 4500 lbs; 1 
= 1000. 
From formula G, we have 

2 X — 

T ~" ~mr * E" * y7 "~ i ^ 3000 ^ "~ 

8 
50 1 

;i = 16| inches ^20- = I4;4- 

From formula 6, section 2, we have mW I = nfhh? 

, r.Wl ^X^^^^X ^'' _ 81 . , 

•'• ^ - "^^ijw - 1 ;; 2500 - "250" "^^^• 

-^ g X 9000 X -g- 



Section 5. — Beams of uniform strength. 

Case 1. — Any beam of uniform strength, and uniform depth, 
not fixed at both ends. 

From the equation of the elastic curve, formula E, section 3, 
we have 



BEAMS OF UNIFORM STRENGTH. 63 

E I — — = M • E — — (1^ 

M = or -^ = — from formula 6, section 2. 

2/0 I 2/0 

Since the depth is constant, y^is constant .-. equation (1) 
becomes 

d'^ V M f 

E -^ = -r =^^ = a constant (2). 

dx^ 1 y^ ^-^ 

ji 4^ = Z ^ + c. 

dx y, 

d y 
Take the origin so that when ic = 0, -^ — = 0, (generally at 

the center- or end of beam) then C = 

f 
.'. ^ y z=z -^ x" + D. When a; = 0, 2/ = .-. D = 0. 

/ 
.'. E 2/ = -^ — x^; - 2/ is a maximum when x ^=z c 

f & . 1 

.-. Vi =■ ir^ — = maximum deflection, c = Z or - ? ac- 

•^ 2 E 2/0 2 

cording to the method of supporting. 

E I 

In section 3, we had the radius of curvature = p = ,^ 

^ M 

As E is a constant, and is constant as above shown for 

this case, then p is constant, or the curve is the arc of a circle. 

Case 2. — Beam of uniform strength, uniform breadth, built 
in at one end, free at the other, and loaded uniformly. 

d'^ y f 

As before, we have E -r4- = 

dx^ y^ 

Let y^ = maximum half depth of beam, and h^ its depth, 
Then M^ =: -^ = nfhh\ 

2 -^ 

c^ __ K _ Vz 1 _ 1 / ^ \ 

(c _ x) "~ ~1F~ ~ y^ '' ~y^ ~ Vz \c — xf 

... E ^ = Zl / 1 \ 

d x^ y^ \ c — X / 



64 BEAMS OF UNIFORM STRENGTH. 

d ij f G 

When a; = -j^ = .-. C = - — log, c 

.'. E y = / logg I If/ X. Integrate by parts. 

Make c — x =i z and integrate. 
E2/=^ j xlog, (-1-)+ clog, {c-x)- (c-a:) }+ F 
When x= 0, 3/ = .-. F =-"—(- c log, c + c) 

JZ 

.-. E 2/ = -^ I ^ log, (^^) + c log, (c - :r) - 
(c — a?) — c log, c -\- c)^ 

= -^ I log, c X (x - c)-ir loge (c - x) X (c - a) + a; j- 

2/ is a maximum when a? = c 

_ f^' - /^' 
^^ ~ Ey^ ■" E2/^ * 

Case 3. — A beam of uniform strength, uniform breadth, 
built in at one end, and loaded at the other. 

Case 4. — A beam of uniform strength and breadth, loaded 
at the middle and supported at the ends. 
2 fc' 
^' " 3E7/ 
Case 5. — A beam of uniform strength and breadth, loaded 
uniformly, and supported at the ends. 



= (f -')'■' 



B Vz 



1 rjQ 1 />• o 

Note.— In case 5 integrate the expression sin — dx by making sin - = 2,or - 
= sin s, and it becomes cz cos zdz. Integrate by parts and substitute the value of z. 



BEAMS OF UNIFORM STRENGTH. 65 

Case G. — A beam of uniform strength, depth, and load, fixed 
at its ends. 

For the plan and elevation of snch a beam, see Figs. 144 
and 143 of A. M. 

Contra-flexure. 

As the beam is of uniform strength and depth, the radius of 
curvature will everywhere be the same. (See Case 1.) And as 
it is fixed at C and A; the point of contra-flexure must be, half 
way between A and C. (Fig. 143 A. M.) 

.-.CB = - = !?. (1.) 

2 4 ^ ' 

Maximum defiection. 

It is obvious that the portion BAB is in the condition of a 
uniformly loaded beam, supported at its ends, whose length is 
c = "I ?.■ Hence, as the beam is of uniform strength and depth, 
its deflection will be found by Case 1. 



/(I)"' 



y! = — — = -^-- — — -A= — = vertical distance between 

A and B. 

By the geometry of the figure, it will be seen that the verti- 
tical distance between C and A, is twice that between A and B. 

Moment of flexure at A. 
Since the portion B A B is similar to a uniformly loaded beam 
supported at the ends, the supporting forces are — , and the mo- 
ment anywhere between B and BisM = — I x J 



(!-)• 



The moment at A is given by making x = 0, 

...M„ = ^=:^=zi. (3.) 

8 16 32 ^ ^ 



66 BEAMS OF UNIFORM STRENGTH. 

Moment at C. 
From B to C the beam is in the condition of one fixed at one 
end, loaded at the other with — , and miiformly with w. Hence 



the greatest moment of flexure is at C, where M^ r= i 1 

c / w c\c _Z loc" _^W c _ZW I 



Comparison of widths at A and C. 

From equations 3 and 4, we see that the moment at C, is three 
times its value A. Hence, as the depth is constant, the 
breadth at the end must theoretically be three times the center 
breadth. 

See theoretical plan of the beam, fig. 144 A. M. 

Moment of flexure anywhere. 
For the general expression for the moment of flexure, we have 

w 
M^ = wc {c — x) — — {c — xY — fl, 

where ^ is the moment required to hold the ends horizontal; 

hence it is equal to the moment at C := '— — — = ~~^~ 

^- . Tvr / \ '^^ / .„ 3 Wc 

see equation 4. .-. M^ z=z wc{c — x) — tt \^ — ^) ^ — 

W / Z^ <, \ 3 W Z 



(t--)- 



2 Z \ 4 / 32 



= ^^{r2-Q- (^•) 



= the general equation of the moment of 
flexure for any point in the beam. 



ARTICLE 365. 67 

Article 365. 

Equation 1. — The following method of deducing the formula 
for the centrifugal force of a unit of mass leads to Equation 1, 
and is a good exercise for the student in the application of some 
fundamental principles of dynamics : 

Let R = the resultant accelerating force acting on a unit of 
mass. Let the path of this particle be referred to three rec- 
tangular axes X, Y, and Z. 

Then since the force is measured by the product of mass and 

d? s 
acceleration, and the mass = 1, we have R = 

d? y 
—T-.Y = component of R along X, 

6} y 

-Y-^ = component of R along Y, 

6} z 

J 3 = component of R along Z, 

••• ^ = \-df-) + \-Tf) + \-d¥) - y-dTf + 

(d^ s \^ / d"^ s \^ 

-jj-J (subtracting and adding \-jj^-J )• 

The square of the radius of curvature of any curve is 



(d s' y 



(See 



Church's Calculus, page 146, Equation 2.) 

•Substituting the value of the denominator of the second 
member of this equation for the same quantity in the value of 

R above, we have R^ = (4^)'^ + (-Jf )^ = (j)' + 

(d^ s\^ 
7 ,2 I * Since R^ = the sum of the squares of two quanti- 
ties, they must be the rectangular components of R; and as 

(d^ s \ 
-rj- 1 is the force in the direction of the mo- 



68 ARTICLES 390-443. 

tion at any instant, the other I — I must be the deviating 
force = centrifugal force. 

Article 390. 

Equations 1 and 2. — Since the radius vector is normal to the 
curve, it coincides in direction with the radius of curvature, 
and both are shifted through the same angle d i in the time d t. 
The arc described by the point A, Fig. 187, A. M., is measured 
by p c? ^ ; or since- c r is the velocity of A, by c r c? t Hence, 
Equation {I) p d i =z c r d t. 

The distance through which the end T of the radius vector is 
carried in the time d t^ owing to the revolution about C, is & r^ d tj 
measured along the arc, whose center is at C. Its projection on 
a perpendicular to AT is br^dt cos 0, which is the arc subtend- 
ing the angle through which r has been shifted negatively. 
This arc divided by r gives the angle 



hrc^dt cos 6 crdt hr^dt cos d 



/ hr^ cos d\ 



Article 402. 

n . . 

Read - instead of n, in the paragraph preceding Eq. (6). 

Article 421, 

-r-, • T^-i^«Qp d . A u p 

Equation 2. — Read — -, — - = = — • 

■ ^ d s d s 

Article 422. 
Equation (1). — The last member should read — — • 

Article 423. 
Equation 2. — The second parenthesis of the first member 

should be (. 4^ + ^ 4^ + ^. 4^ + 4^1 
\ d X ~ dy ~ d z ^ dt F 

Article 443. 
The eccentricity of an elli2)se is the distance from one focus to 

the center, divided by the transverse axis, or — — . 



ARTICLES 444-462. 69 

Article 444. 

X Y = V(GH' 4- XW). A section of the hyperboloid 
througli X Y, and perpendicular to A B, Fig. 195, A. M., would 
be a circle of radius X Y, and tlie projection of the generating 
line in the position F E, on the plane of this section would be a 
vertical line at a distance G H from the center of the section. 
The point "W is where this line cuts the circle, and its distance 
from the axis of the paraboloid is evidently 

X Y = (GH' + XWy. Let the student draw the section as 
above directed. 

Article 450. 

Equation (1). The velocity of Pj in a plane perpendicular to 
Ci is a, Ci Pj. Since the line of action makes the angle i with 
the axis Cj, it makes (90° — i^) with the direction of the motion 
of Pj. The component of the velocity of Pj along the line of 
action, will, therefore, be a^.Q^ Pj cos (90° — i^ = a^ Cj P, sin 
«i. By the same reasoning the velocity of Pg along the line of 
action is a^ Cs Pg sin i^, and these are equal by the principle stated 
in this article. Hence Equation (1). 

Article 458. 

Equation (1), r is the length of cord unwound from the circle 
whose radius is Cj Pj (Fig. 198), while a distance q on the circle 
whose radius is C, I passes the point I. These arcs are propor- 
tional to the radii of their respective circles, or 

^ 0, P, 
— = -^ = sm 0. 
q Ci I 

Article 462. 
Equation (3). Substituting values from Equations (1) and (2) 
of this article in the equation 5 = 2 I 1- — 1 / rdqwe 

/I 1\ /'^(-f-e) q 

have s = 4.r,y--^—JJ^ sm -^ d q. Multiply 

and divide the second member by 2 r^, and we have s = S Tq 



70 ARTICLES 463-483. 

(_Cos(|--e) + l) = 8i (l + l)(l_smfl). 

Article 463. 
See Willis- Principles of Mechanism, page 137. 

Article 482. 
The degree of approximation attained by putting 



Vc^ — {;,\ — r;)^ =-c — ^ ' ^ ^ — is seen by squaring both 

members of the inequality. The term ^ is neglected in 

the second member. 

Article 483. 
Equation (1). — The first value oi y = r^ — ^ T" ^ is obvi- 
ous from an inspection of the figure of the conoid. To obtain 
the equation r^ — -^—^ — - = -^-^ — it is to be considered 

that the length of belt for two equal pulleys of radius r^, is L = 
2 c -\- 2 TT r^, which, placed equal to the approximate value of 
L from Equation (3 A), Article 482, gives 2c-|-27rro=:2c 

+ ^ {r, -^ r,) + ^' '^ OTT, + r, = 2 r, - ^' /^ 

O TT C 

which is also Equation (2) of this article. The manner of using 
this equation for designing a pair of stepped cones in which 
two opposite pulleys are of equalradii is explained in the A. M. 

If it is required to complete a pair of cones when the radii of 
a pair of opposite pulleys are given, the value of r^ may first be 
found by substituting the given values for r^ and r^, in equa- 
tion (2). 

Then replacing r^ by this value, the radius of one of the re- 
quired pulleys may be assumed and the radius of its opposite 
pulley calculated by solving the resulting quadratic equation. 

Problem. — In a pair of stepped cones let the radii of the end 
pulleys opposite each other be r^ = 24'', r., = 6'', and let the 
distance of their centers be c = 48'' required the radii of the 



ARTICLES 488-491. 71 

adjacent pair of pulleys on the stepped cones. From (2) we have 
2 r, = (r, + r,) + tl^' = 30 + 3^^^ (^ = 3^ nearly) 

= 32.15 nearly. .-. r, -^ r,= 32.15 - ^^^ ~ ^^^^ . 

TT C 

Let rj and r^ now represent respectively the radii of the 
greater and smaller of the pulleys to be next calculated. 

We may now fix the value of r^ and calculate r^. It will how- 
ever generally be easier to find r^ by trial than to solve the 
quadratic equation. 

Let Ti be fixed at 20'', and for the first trial let us suppose r^ 
to be equal to 11.8". Then r^ -{-^2 = 31.8, r^ — r^=: 8.2'', and 

we have from the formula 31.8 = 32.15 — ^^— = 31.7 

1056 
nearly. 

Again let r^ = 11.6" for a second trial, then 31.6 =32.15 
— .468 = 31.68. Thus we see that the true value of r^ lies be- 
tween 11.6" and 11.8"; r^ = 11.7" very nearly satisfies the 
equation. 

Article 488. 

In Fig. 218 let C^ T^ = r, T, % = I, the angle T^ C^ T^ = &, 
and the angle T2 T^ C2 = /3. Then C2 A = (r cos a + ^ cos (3) 
tan B. 

Placing the above value of C2 A in equation (2), and elimina- 
ting /3 by means of the equation, I sin fi = r sin a, we have Vj_ 



== ^2 sm a ■< 



J cos a 



y L— — sin^a 



+ 1 



> from which the velocity of 



the piston of an engine may be calculated for any given crank 
angle ; for by knowing the number of revolutions, the value of 
v^ is easily found. 

Article 491. 

Equations (2) and (1). — Let the axis C2 be perpendicular to the 
plane of the paper and let the plane of the axes of the shafts be 
vertical. The projection on the paper, of the path of Fg will be 
the circle F^ ¥^ B D, and the projection of the path of F^ will be 
the ellipse F^ H B E of the accompanying figure. 



72 



AETICLES 488-491. 




Suppose F2 to move to the position F/, Fj will in the same 
time move to F/. The angle F^ F/ is cp,. The angle F^ F/ 
which is also equal to 62, is the projection of the angle de- 
scribed by Fi in the plane of its path. The true magnitude 
of this angle is F^ A, (found by revolving the plane of the 
ellipse in which the angle lies, about F^ B until the elUpse coin- 
cides with the circle, when F^ will fall at A in the vertical line 
M A.) The angle P^ A is ^j. (See the A. M.) 

Then M cot (^^ = M A. M tan (^2 = M F/. 

tan 02 M F/ H . ^ 

.-. i-i- == = 7^-=- = cos t or tan 0, tan 6^ = cos ^. 

cot 01 MA F2 ^' ^' 

Differentiating this equation we have 

c^ 02 t^ii 02 cos^ 02 tan 02 (1 -j- tatf 0^) 

d (t>, t£ 



cot ( 



1 + 




tan 01 (1 -f- tan^ 02) 
tan 01 -}- cot 01 



cot 05 



tan 02 = 



\ COt^ 02 / 

tan 01 -{- cot 01 
tan 02 -\- cot 02 
cos i 



tan 02 + cot 02 . * 



Eliminate 02 by Equation (1), 



tan 01 

cos ^ (1 + tan^ 0i) 



Then ^ - , 2 , 2 • 
«! tan'' 01 -j- cos^ I 

When Fi is in the plane of the axes 0i = 0, tan 0i = and 
— = . which is the first of Equations (1) A. M. 



In a simi- 



ARTICLE 507. 



73 



lar manner by eliminating 0i, the second of Equations (1) may 
cos ?' (1 4- t^^^ 01 ) 



be obtained. When — = 



1 or the 



tan^ <^i -|- cos^ ^ 

shafts have equal angular velocities, we find tan^ ^^ = cos t; 0i 
= tan-^ Vcos I. (See Willis' Principles of Mechanism, Article 
513.) 

Article 507. 

Equations (3). — The square of the perpendicular, drawn from 
any point in the circumference of a circle to a diameter, is equal 
to the product of the segments into which the foot of the per- 
pendicular divides the diameter .*. I --) = T V (2 r — T Y) 
= T Y . 2 C M. 

An exact parallel motion^ accomplished entirely by means of 
link work (without any sliding piece, as in Watts' Exact Parallel 




Motion, Article ^07, A. M.), has recently been discovered by 
M. Peaucillier, an outline oi which is here given. 



7-i AETICLE 507. 

In tlie accompanying figure, F E G is an equilateral paral- 
lelogram of bars, jointed at each of its corners. A and C are 
fixed points; A F and A G are links jointed at their extremi- 
ties. C E is also a link jointed to the corner (E) of the paralel- 
ogram, and capable of turning about C. When the point is 
moved in the plane of the figure, the points F and G must 
describe arcs of circles about A. E must describe an arc 
about C. 

When this combination of links is made to oscillate about the 
fixed points A and C : 

1st. If C is midway between A and E, the point will be 
guided in a straight line perpendicular to A 0. 

2d. If C is in any other position on the line A E, the point 
will move in the arc of a circle. 

To prove these propositions, and also to find exact relations 
between the radius of the arc described by 0. and the position 
of C on the line A E, is the object of the following analytical 
investigation : 

Suppose the instrument to be swung aside until the axis A 
takes the position A B, making an angle d with its neutral posi- 
tion. E will be at E' on this line, and also on the circle whose 
center is at C. The parallelogram will be partially closed as 
represented by the dotted lines in the figure. will have 
moved to some point B on the new position of the axis, and it 
is required to find the locus of this point. 

Let be the origin, X the horizontal, and Y the vertical 
axis. Then M = :r, M B = ?/. 

'LetA¥ = AG = l; F E = E G = 5; A E' = c; A = A:; 

C E = r ; C A = 6?, and the angle F' E' B = F' B E' =: /3. 

Then x =z {k — y) tan (1) 

y = Jc — (c -[- 2 s cos /3) cos (2) 

Z2 = c'^ -f- 52 -f- 2 c 5 cos /3 (3) 

A E' = c = c? cos + Vr' - d' sin^ d (4) 

Combining 2 and 3 we have 

(p ^2 ^.2V ^2 ^9 
c -1 I COS = ^ cos d (5) 



Combining this with (4) 

_ (P - s') CO 

~~ ~d COS d -\- y^/r^~Z 
Putting d in terms of tan d gives 



y^,—, (l-^)<^°'' (6) 

dcosd -{- ^r' — d'' sin^ 



ARTICLE 515. 15 

P — 52 



y =: fc ^ 







v; 


I + tan^ 


e 






d 




- 52 


- 


d' tan^ d 


VT 


4- 


tan^ 

Z2 - 


1 + tan^ 



__ y^ 

c? -I- Vr^ + (r^ — d') tan^ 

Substituting 6 from (1), transposing Z; and changing the signs 
of both members 

^_ ^ (/^ - y) (P - .^) 

y ~ d{k — y) ^ ^r' {k - yf + x' {r' - d') ' 

Dropping the factor (k — y), clearing of radicals, and placing 
(P — s') = n, we have (r^ — d') {k — yf + (r^ _ d') x^ — ri" — 
2nd(k -y) (7) (r^ - d') ¥ - 2k (/-^ - d') y + (r^ - c^^^ 7/^ -f 
(r^ _ (^2^ a:;2 = n^ — 2 n d k -\- 2 n d y, which is apparently 
the equation of an ellipse (since S^ — 4 a c < 0. See Church's 
Analytical Geometry, page 207). 

Since the center of this curve is evidently on the line A 
produced, which contains one of the principal axes, let us 
transfer the origin of coordmates to some point on this line at a 
distance a from 0. The formulae for such transformation are 
xz=zx'] y = y' ^ a. Then from (7) (r^ _ d') {k — a — yj -j- 
{r'' - d'):xf^ =z 'n? -2nd{k - a — y'). 

Let k — a =1 h, and drop the primes; {r'^ — d^) x^ -f (r^ —^d^) 
y^ = n^ — 2ndb — {r' — d')b-'J^ ^ 2 n d -\- 2 (r^ — d"") b )■ y. 
If the origin is at the center of the curve, the term containing y 

must disappear or 2 n d -\- 2 (r^ — c?^) J — ; b =: j^ • 

Eeplacing b by its value, we have k — a =z — — — 

n d 

Since the co-efficients of x"^ and y"^ are equal, the ellipse has 

equal axis, or is a circle of radius a = k — - • When d 

. r^ — d^ 

= r^ a= oc, or the locus is a straight line Q . E . D . 

Article 515. 

Problem. — Find the formula for the work performed upon 
the piston of a steam engine during one stroke, the length of 



76 AETICLE 531. 

stroke being ?, the area of the piston a, the initial pressure of 
the steam p^, the steam being cut off at a distance c from the 
beginning of the stroke. 



Ans. ap,c {l -f-log,— )• 



Article 531. 

Equation (1). — Let / = the force of gravity at any point on 
the earth's surface at the level of the sea. 

Let T = time of one revolution of the earth on its axis. 

Let n = number of revolutions in a unit of time. 

Then nT = I. 

The centrifugal force of a unit of mass at the level of the sea 

and latitude X is 

^2 (2 ,r . R cos X.nY 4 tt' R 

F = = -^ = cos X. 

R cos X R cos X T^ 

The component of F acting opposite to gravity is / = F cos X 

Gravity being diminished by this amount, if we subtract it 
from G the real attraction of the earth, we shall have 

2 TT^ R 2 TT^ R 

/ = Gl- ^— ^— cos . 2 X. 

The value of /for latitude 45° has been found by experiment. 
Denote this value by g^. 

Then from the formula (making X = 45°) 

2 tt' R , 2 TT^ R ,_. 

9i=^^ p-- . ••• ^ = ^1 T^-cos (2)- 

For a point at a distance h above the earth's surface (h being 
small compared with R) we have from Newton's law of gravity 



m 



\R4-A/ 



, R^ , R^ 



(R-l-A)'^ ^ R^-f 2R;i + /i''^ 



Performing the division to two places 

- ^0 -4^)-= -('-S?" ")('-!> 



articles 537-557. 77 

Article 537. 

Equation (2) — Let a be the angle whicli a tangent to the curve 
makes at any point. 

W W 

Then Q = ^<; cos a -. 



J I + tan^ a i/i _l i^ 
^ ^ ^ (dxf 



Since the horizontal component of the velocity of the projec- 
tile is constant we have 

t'o cos 
Vn cos d = V cos a. .-. cos a == 

V 

Vq cos 6 



.-. Q =W. 



Article 545. 



Equation (1) — The component of g urging the pendulum 
towards its lowest point is g sin d in which is the inclination 
of the curve to the horizon. To show that in the cycloidal pen- 
dulum this force is proportional to the length of the arc between 
any position of the pendulum and its lowest position, we con- 
sider that the length of any curve is 5= / d s =' / p d d. 
From Art. 390 A. M., we have for the cycloid p = 4 r.^ cos 0. 
.-. 5 =: 4 Tg / cos c? = 4 rg sin which varies as sin 0. 

fj o 

Hence the vibrations of a cycloidal pendulum are isochrorxous. 

Article 552. 

Problem. — A body weighing 100 lbs. starts from a point 48 
feet above the earth's surface. After falling freely 16 feet it 
encounters a constant resistance equal to | gravity. Required 
its potential energy before starting ; its actual energy, when at & 
distance of 16 feet from the earth ; its potential energy at the 
same point ; the sum of its actual and potential energies when 
it reaches the earth ; also the work done in the fall. 

Article 557. 
Equation 6 should read as follows : 

— V- — — = -— — sm^ at H- - — ! cos^ at = -^i— i . 

2^^2 2 g ^2 2 



78 



ARTICLE 578. 



Article 578. 

A method of finding the moment of inertia of solids of revo- 
lution is illustrated in the solution of the following problem. Find 
the moment of inertia of a cone whose altitude is h and radius 
of base r, about an axis through 
its vertex and perpendicular to 
its geometric axis. Conceive a 
portion of the cone included be- 
tween two planes at right angles 
to its axis, and at a distance dx 
apart. Find the moment of inertia 
of this elementary slice or disc, by 
applying to it the theorem of Art. 
576 A. M. 




This will give - 



■ dx . -\- w TT y^ dx . x^. 



,\\= w - j y^ dx J^ lu TV I 



y'^ x^ di 



20 



. (r^ -f 4 W) 



If the moment of inertia about the axis X were required, 
the axis of inertia would pass through the center of gravity of 
the disc and be perpendicular to its bases. We should there- 
fore have I3. = — - / y^ d x. This is the formula for problem 
2 t/ o 

I of this Article. 

Problem III. — Although the ellipsoid is not a solid of revo- 
lution, its moment of inertia may be found by the method above 
explained. 

Let the origin be at the center of the ellipsoid and let X, 
Y and OZ be in the direction of the semi-axes «, h and c re- 
spectively. A section perpendicular to X is an ellipse whose 
semi-axes are y and z. The moment of inertia of the elemen- 
tary slice with respect to X is therefore 



w TT y z 



4 
.-.I 



dx -j- 



w IT y z^ 



- d X. (See problem in Notes on Art. 95.) 






zdx-\ —J _JJ i 



dx. 



4: J -a"" '4 

The values of z and y are taken from the following equations, 
j a2 z" -I- c^ a;2 = a^ & 



ARTICLES 579-590. 79 

.-. I = t. ^' f \a^ - xj dx + '—^ f ^dx id' - x^) 
4, cc J -a^ ' 4: a- t/ -a ^ ' 

4: ^u TT a h c (h^ -\- c^) 
— _ 

Let tlie student perform the same problem by integrating the 
following formula according to the directions in Note on Art. 83, 
assigning the proper limits. 

\ = IV I I I dx dy dz (t/^ -|- z'). 

Theorem. — The actual energy of a body revolving about a 
fixed axis is found by multiplying its moment of inertia by the 
square of its angular velocity and dividing the product by twice 
the force of gravity. 

Let p = the radius of gyration of the body, a its angular ve- 
locity, and W its weight. If the mass of the body were con- 
centrated at the center of gyration, (that is, at a distance p from 
the axis of rotation) its velocity would be a p. Its actual energy 

would be—- a" 0^=-^. W p2 = — - L 

If the axis of rotation traverses the center of gravity of the 
body and is a " line of symmetry of the figure of the body," the 
values of 1 and p may be taken from the columns headed I^, and 
Po in the table of Art. 578 A. M. (See Art. 589 A. M.) 

Aeticle 579. 

Read instead of under example. 

6 6 • ^ 

Article 590. 

Equation 5. — From equation ((3) of Art. 585, we haves^=:-. 

n I 

From equation (6) of Art. 588 we have cos : 





I 


5 v^r+K^ 


n V I' -\- K.^ 

Or 1 =r.-^\r-f K^). 
/3 + II cos^ y. ,, 


1 n-" (V 4- K^) 
I 1^ 

= I'^ + K-^ = I^ cos2 « -f l:] cos^ 

Equation (2) 



Note.— The moment of inertia as here used is defined in Arte. 571, 572, and 573, A. M. 



80 ARTICLES 606-648. 



But I = Ij cos^ a 4- I2 cos^ /3 + I3 cos^ y. (Eq. (3), Art. 585) 

II COS^ X , I2 cos'' j3 , I3 COS^ y ^. „ , ^. o ^ , T9 

71^ n^ 91^ 

COS^ y. 

Or (i; - ^) cos^ a + (11 - ^) cos^ ^ + (i| - ^,) 

COS^ y = 0. 

Article 606. 

First Equation. — The moment of the deviating couple in 
terms of E is M = 2 E tan a. The value of the constant B is 
the sum of its components with reference to the axes C G and 

a^ I a^ I 

C E, and these components are — — ^ cos^ a and -r — - cos^ 

2 ^ 2 ^ 

(180 - (90° _ a)) = - ^ sin^ a .-. M = j (I, - I,) 

W 
cos a sin a =:: — a^ (p^^ — p^) cos a sin a. 

Article 638. 

Formula for computing the discharge and diameter of pipes, 
are given in Article 450 of Rankine's Civil Engineering. 

Article 648. 

Equations (1), (2), and (3). — These equations may be obtained 
as follov/s : The velocity of the jet before it meets the surface 
is V. V/hen the surface has changed its direction through an 
angle /3, the component* of its velocity in the original direction 
of the jet is v cos /3. The change of velocity is, therefore, v — 
V cos /3 == V (1 — cos /3). Hence, the change of momentum 

v\^hich measures the force in this direction, is F_ r= v 

9 

(1 — cos /3). The change of velocity in a direction perpendicu- 
lar the original direction of the jet is v sin /3 .-. F^ ^ 

a9_ V sin /3. The resultant F = Vp/ 4. F/ = JL^ 
9 O 

Vn - cos /3)'^ 4- sin" /3 =zt9L y V2 (1 — cos/3)=:-^ v 



.•0/3 2pQ . /3 

2 ij 2 



1 



